90. Mathematical Induction Solutions Part 1

  1. Let \(P(n) =1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}\)

    \(P(1) = 1^2 = \frac{1.2.3}{6} = 1\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n= m.\) For \(n = m + 1\)

    \(P(m + 1) = 1^2 + 2^2 + \ldots + (m + 1)^2\)

    \(= \frac{m(m + 1)(2m + 1)}{6} + (m + 1)^2\)

    \(= \frac{(m +1)(2m^2 + m + 6m + 6)}{6}\)

    \(= \frac{(m + 1)(2m^2 + 3m + 4m + 6)}{6}\)

    \(= \frac{(m + 1)(m+ 2)(2m + 3)}{6}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  2. Let \(P(n) = \frac{1}{1.2} + \frac{1}{2.3} + \ldots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\)

    \(P(1) = \frac{1}{1.2} = \frac{1}{1.(1 + 1)}\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) = \frac{1}{1.2} + \frac{1}{2.3} + \ldots + \frac{1}{m(m + 1)} + \frac{1}{(m + 1)(m + 2)}\)

    \(= \frac{m}{m + 1} + \frac{1}{(m + 1)(m + 2)}\)

    \(= \frac{m(m + 2) + 1}{(m + 1)(m + 2)} = \frac{m+ 1}{m + 2}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  3. Let \(P(n) = 1^3 + 2^3 + \ldots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2\)

    \(P(1) = 1^3 = \left(\frac{1.2}{2}\right)^2 = 1\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n= m + 1\)

    \(P(m + 1) = 1^3 + 2^3 + \ldots + (m + 1)^3\)

    \(= \left(\frac{m(m + 1)}{2}\right)^2 + (m + 1)^3\)

    \(= \frac{(m + 1)^2(m^2 + 4m + 4)}{4}\)

    \(= \left(\frac{(m + 1)(m + 2)}{2}\right)^2\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  4. Let \(P(n) = 1.3 + 2.3^2 + \ldots + n.3^n = \frac{(2n - 1)3^{n + 1} + 3}{4}\)

    \(P(1) = 2 = \frac{1.3^2 + 3}{4} = 3\)

    Thus, \(P(n)\) is true for \(n= 1.\) Assume that it is true for \(n = m.\) For \(n= m + 1\)

    \(P(m + 1) = \frac{(2m - 1)3^{m + 1} + 3}{4} + (m + 1)3^{m + 1}\)

    \(= \frac{3^{m + 1}(2m - 1 + 4(m + 1) + 3)}{4}\)

    \(= \frac{(2m + 1)3^{m + 2} + 3}{4}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  5. Let \(P(n) = \cos\alpha + \cos 2\alpha + \ldots + \cos n\alpha = \sin \frac{n\alpha}{2}\text{cosec}\frac{\alpha}{2}\cos\frac{(n + 1)\alpha}{2}\)

    \(P(1) = \cos\alpha = \sin\frac{\alpha}{2} \frac{1}{\sin\frac{\alpha}{2}} \cos\alpha\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n= m + 1\)

    \(P(m + 1) = \cos\alpha + \cos 2\alpha + \ldots + \cos (m + 1)\alpha = \sin \frac{(m + 1)\alpha}{2}\text{cosec}\frac{\alpha}{2}\cos\frac{(m + 2)\alpha}{2}\)

    \(= \sin \frac{m\alpha}{2}\text{cosec}\frac{\alpha}{2} \cos\frac{(m + 1)\alpha}{2} + cos(m + 1)\alpha\)

    \(=\text{cosec} \frac{\alpha}{2} \left[\sin\frac{m\alpha}{2} \cos\frac{(m + 1)\alpha}{2} + \cos(m + 1)\alpha \sin\frac{\alpha}{2}\right]\)

    \(= \frac{1}{2}\text{cosec} \frac{\alpha}{2} \left[ \sin \frac{(2m + 1)\alpha}{2} - \sin\frac{\alpha}{2} + \sin\frac{(2m+ 3)\alpha}{2} -\sin \frac{(2m + 1)\alpha}{2}\right]\)

    \(= \frac{1}{2} \text{cosec} \frac{\alpha}{2} \left[2\cos\frac{(2m + 3 + 2)\alpha}{2}\sin\frac{(2m + 2)\alpha}{4}\right]\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  6. Let \(P(n) = \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \ldots + \tan^{-1}\frac{1}{n^2 + n + 1} = \tan^{-1}\frac{n}{n + 2}\)

    \(P(1) = \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{1}{1 + 2}\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n= m + 1\)

    \(P(m + 1) = \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \ldots + \tan^{-1}\frac{1}{m^2 + m + 1} + \tan^{-1}\frac{1}{(m + 1)^2 + m + 1 + 1} = tan^{-1}\frac{m + 1}{m + 3}\)

    \(= \tan^{-1}\frac{m}{m + 1} + \tan^{-1}\frac{1}{m^2 + 3m + 3}\)

    \(= \tan^{-1}\left[\frac{\frac{m}{m + 2} + \frac{1}{m^2 + 3m + 3}}{1 - \frac{m}{m + 2}\frac{1}{m^2 + 3m + 3}}\right]\)

    \(= \tan^{-1}\left(\frac{m + 1}{m + 3}\right)\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  7. Let \(P(n) = {}^nC_1 + 2.{}^nC_2 + \ldots + n.{}^nC_n = n.2^{n - 1}\)

    \(P(1) = {}^1C_1 = 1 = 1.2^{1 - 1} = 0\)

    Thus, \(P(1n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n= m + 1\)

    \(P(m + 1) = {}^{m + 1}C_1 + 2.{}^{m + 1}C_2 + \ldots + (m + 1){}^{m + 1}C_{m + 1} = (m + 1)2^m\)

    \(= ({}^mC_0 + {}^mC_1) + 2({}^mC_1 + {}^mC_2) + \ldots + m({}^mC_{m - 1} + {}^mC_m) + (m + 1)({}^mC_m + 0)\)

    \(= ({}^mC_0 + 2.{}^mC_1 + \ldots + m^{}mC_{m - 1} + (m + 1)/{}^mC_m) + ({}^mC_1 + 2.{}^mC_2 + \ldots + m.{}^mC_m)\)

    \(= ({}^mC_0 + 2.{}^mC_1 + \ldots + m^{}mC_{m - 1} + (m + 1)/{}^mC_m) + m.2^{m - 1}\)

    \(=2^m + m.2^{m - 1} + m.2^{m - 1} = (m + 1)2^m\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  8. Let \(P(n) = u_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 +- \sqrt{5}}{2}\right)^n\right]\)

    \(P(1) = u_1 = 1 = \frac{1}{\sqrt{5}}\left[\left(\frac{1 + \sqrt{5}}{2}\right) - \left(\frac{1 - \sqrt{5}}{2}\right)\right] = 1\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) and \(n = m + 1\) For \(n= m + 2\)

    \(P(m + 2) = u_{m + 2} = u_{m} + u_{m + 1}\)

    \(= \frac{1}{\sqrt{5}} \left[\left(\frac{1 + \sqrt{5}}{2}\right)^m \left(\frac{1 + \sqrt{5}}{2} + 1\right) - \left(\frac{1 - \sqrt{5}}{2}\right)^m \left(\frac{1 - \sqrt{5}}{2} + 1\right)\right]\)

    \(= \frac{1}{\sqrt{5}}\left[\left(\frac{1 + \sqrt{5}}{2}\right)^{m + 2} - \left(\frac{1 - \sqrt{5}}{2}\right)^{m + 2}\right]\)

    Thus, \(P(n)\) is true for \(n = m + 2.\) Henece, we have proven the equation by mathematical induction.

  9. Let \(P(n) = 11^{n + 2} + 12^{2n + 1}\)

    \(P(1) = 11^3 + 12^3 = (11 + 12)(121 - 11*12 + 144) = 23 * 233\) which is divisible by \(133\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = 11^{m + 3} + 12^{2m + 3}\)

    \(= 11.11^{m + 2} + 144.12^{2m + 1}\)

    Dividing \(P(m + 1)\) by \(P(m)\)

    \(P(m + 1) = 11.P(m) + 133.12^{2m + 1} = 11.133k + 133.12^{2m + 1}\) where \(k\) is an inetger.

    \(P(m + 1)\) is divisible by \(133\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  10. Let \(P(n) = p^{n + 1} + (p + 1)^{2n - 1}\)

    \(P(1) = p^2 + (p + 1)^{2.1 - 1} = p^2 + p + 1\) which is divisible by \(p^2 + p + 1\)

Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

\(P(m + 1) = p^{m + 2} + (p + 1)^{2m + 1}\)

Dividing \(P(m + 1)\) by \(P(m),\) we get

\(P(m + 1) = p.P(m) + (p^2 + p + 1)p^{2m - 1}\)

Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  1. Let \(P(n) = 2^n > 2n + 1,~\forall~n>2\)

    \(P(3) = 2^3 > 2*3 + 1\)

    Thus, \(P(n)\) is true for \(n = 3.\) Assume that it is true for \(n = m~\forall~m\geq 3\) For \(n= m + 1\)

    \(P(m + 1) = 2^{m + 1} > 2m + 3 = \alpha\)

    Multiplying \(P(m)\) with \(2\), we get

    \(2^{m + 1} > 4m + 2 = \beta\)

    Subtracting the two formulas we have

    \(\beta - \alpha = 4m + 2 - 2m - 3 = 2m - 1 > 0\)

    \(\therefore \beta > \alpha\)

    \(2^{m + 1} > \alpha\) i.e. \(2^{m + 1} > 2m + 3\)

Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  1. Let \(P(n) = n^4 < 10^n,~\forall~n \geq 2\)

    \(P(2) = 2^4 < 10^2\)

    Thus, \(P(n)\) is true for \(n = 2.\) Assume that it is true for \(n = m~\forall~m\geq 2\) For \(n= m + 1\)

    \(P(m + 1) = (m + 1)^4 < 10^{m + 1} = \alpha\)

    \(10^{m + 1} > 10^m = \beta\)

    \(\frac{\beta}{\alpha} = \frac{10m^4}{(m + 1)^4}\)

    \(\because m\geq 2 \therefore m + 1\geq 3\Rightarrow \frac{1}{m + 1} \leq \frac{1}{2}\)

    \(\frac{m}{m + 1}\geq \frac{2}{3}\)

    \(\Rightarrow \left(\frac{m}{m + 1}\right)^4 \geq \left(\frac{2}{3}\right)^4\)

    \(10\left(\frac{m}{m + 1}\right)^4 \geq 10\frac{16}{81} > 1\)

    \(\frac{\beta}{\alpha} > 1 \Rightarrow \beta > \alpha\)

    \(\Rightarrow 10^{m + 1}> \alpha \Rightarrow 10^{m + 1} > (m + 1)^4\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  2. Let \(P(n) = 1^3 + 3^3 + \ldots + (2n - 1)^3 = n^2(2n^2 - 1)\)

    \(P(1) = 1 = 1^2(2.1^2 - 1) = 1\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = 1^3 + 3^3 + \ldots + (2m - 1)^3 + (2m + 1)^3\)

    \(= m^2(2m^2 - 1) + (2m + 1)^3\)

    \(= 2m^4 - m^2 + 8m^3 + 12m^2 + 6m + 1\)

    \(= 2m^4 + 8m^3 + 11m^2 + 6m + 1\)

    \(= (m + 1)^2[2(m + 1)^2 - 1]\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  3. Let \(P(n) = 3.2^2 + 3^2.2^3 + \ldots + 3^n.2^{n + 1} = \frac{12}{5}(6^n - 1)\)

    \(P(1) = 3.2^2 = 12 = \frac{12}{5}(6 - 1) = 12\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m~\forall~m\geq 3\) For \(n= m + 1\)

    \(P(m + 1) = 3.2^2 + 3^2.2^3 + \ldots + 3^m.2^{m + 1} + 3^{m + 1}2^{m + 2}\)

    \(= \frac{12}{5}(6^m - 1) + 3^{m + 1}2^{m + 2}\)

    \(= \frac{12}{5}(6^m - 1) + 12.3^m2^m\)

    \(= \frac{12}{5}[6^m - 1 + 5.6^m]\)

    \(= \frac{12}{6}(6^{m + 1} - 1)\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  4. Let \(P(n) = \frac{1}{1.4} + \frac{1}{4.7} + \ldots + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\)

    \(P(1) = \frac{1}{4} = \frac{1}{3 + 1} = \frac{1}{4}\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = \frac{1}{1.4} + \frac{1}{4.7} + \ldots + \frac{1}{(3m - 2)(3m + 1)} + \frac{1}{(3m + 1)(3m + 4)}\)

    \(= \frac{m}{3m + 1} + \frac{1}{(3m + 1)(3m + 4)}\)

    \(= \frac{m(3m + 4) + 1}{(3m + 1)(3m + 4)}\)

    \(= \frac{(3m + 1)(m + 1)}{(3m + 1)(3m + 4)}\)

    \(= \frac{m + 1}{3m + 4}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  5. Let \(P(n) = (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta\)

    \(P(1) = (\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = (\cos\theta + i\sin\theta)^{m + 1}\)

    Substituting the value of \(P(m)\)

    \(P(m + 1) = (\cos m\theta + i\sin m\theta)(\cos\theta + i\sin\theta)\)

    \(=\cos m\theta\cos\theta - \sin m\theta\sin\theta + i(\sin m\theta\cos\theta + \cos m\theta\sin\theta)\)

    \(= \cos(m + 1)\theta + i\sin(m + 1)\theta\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  6. Let \(P(n) = \cos\theta.\cos 2\theta \ldots \cos 2^{n - 1}\theta = \frac{\sin 2^n\theta}{2^n\sin\theta}\)

    \(P(1) = \cos\theta = \frac{\sin 2\theta}{2\sin\theta} = \cos\theta\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = \cos\theta.\cos 2\theta \ldots \cos 2^{m - 1}\theta.\cos 2^m\theta\)

    Substituting the value of \(P(m)\)

    \(P(m + 1) = \frac{\sin 2^m\theta}{2^m\sin\theta}\cos 2^m\theta\)

    \(= \frac{2\sin 2^m\theta\cos 2^m\theta}{2^{m + 1}\sin\theta}\)

    \(= \frac{\sin 2^{m + 1}\theta}{2^{m + 1}\sin\theta}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  7. Let \(P(n) = \sin\alpha + \sin 2\alpha + \ldots + \sin n\alpha = \frac{\sin \frac{n\alpha}{2}}{\sin \frac{\alpha}{2}}\sin \frac{n + 1}{2}\alpha\)

    \(P(1) = \sin\alpha = \frac{\sin\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}\sin\frac{1 + 1}{2}\alpha = \sin\alpha\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = \sin\alpha + \sin 2\alpha + \ldots + \sin m\alpha + \sin(m + 1)\alpha\)

    Substituting the value of \(P(m)\)

    \(P(m + 1) = \frac{\sin \frac{m\alpha}{2}}{\sin \frac{\alpha}{2}}\sin \frac{m + 1}{2}\alpha + \sin(m + 1)\alpha\)

    \(= \frac{1}{2\sin\frac{\alpha}{2}}\left[2\sin\frac{m\alpha}{2} \sin\frac{(m + 1)\alpha}{2} + 2\sin(m + 1)\alpha \sin\frac{\alpha}{2}\right]\)

    We will use the idenity \(2\sin A\sin B = \cos(A - B) - \cos(A + B)\)

    \(= \frac{1}{2\sin\frac{\alpha}{2}}\left[\cos\frac{\alpha}{2} - \cos\frac{(2m + 1)\alpha}{2} + \cos\frac{(2m + 1)\alpha}{2} - \cos \frac{(2m + 3)\alpha}{2}\right]\)

    \(= \frac{1}{2\sin\frac{\alpha}{2}}\left[\cos\frac{\alpha}{2} - \cos \frac{(2m + 3)\alpha}{2}\right]\)

    Using the same identity we used earlier

    \(= \frac{1}{2\sin\frac{\alpha}{2}}2\sin\frac{(m + 1)\alpha}{2} \sin\frac{(m + 1)\alpha}{2}\)

    \(= \frac{\sin\frac{(m + 1)\alpha}{2} \sin \frac{(m + 1)\alpha}{2}} {\sin\frac{\alpha}{2}}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  8. Let \(P(n + 1) = a_{n + 1} = \frac{1}{(n + 1)!}\)

    \(P(1) = \frac{1}{1!} = 1\)

    \(P(2) = a_2 = \frac{a_1}{1 + 1} = \frac{1}{2} = \frac{1}{2!}\)

    Assume that \(P(m)\) is true i.e. \(P(m) = \frac{1}{(m - 1)!}.\) For \(n = m + 1\)

    \(a_{m + 1} = \frac{a_m}{(m + 1)}\)

    Substituting the value of \(P(m)\)

    \(a_{m + 1} = \frac{1}{m!(m + 1)} = \frac{1}{(m + 1)!}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  9. Let \(P(n + 2) = a_{n + 2} = 5a_{n + 1} - 6a_n, n\geq 1\)

    For \(n = 3\)

    \(P(3) = 5a_2 - 6a_1 = 25 - 6 = 19 = 3^3 - 2^3 = 19\)

    Assume that \(P(m)\) and \(P(m + 1)\) is true i.e. \(P(m) = 3^m - 2^m, P(m + 1) = 3^{m + 1} - 2^{m + 1}\)

    For \(n = m + 2.\)

    \(a_{m + 1} = 5a_{m + 1} - 6a_m\)

    \(= 5(3^{m + 1} - 2^{m + 1}) - 6{3^m - 2^m}\)

    \(= 5(3^{m + 1} - 2^{m + 1}) - 2.3^{m + 1} + 3.2^{m + 1}\)

    \(= 3.3^{m + 1} - 2.2^{m + 1}\)

    \(= 3^{m + 2} - 2^{m + 2}\)

    Thus, \(P(n)\) is true for \(n = m + 2.\) Henece, we have proven the equation by mathematical induction.

  10. Let \(P(n) = u_n = 2^n + 1, n\in N\)

    For \(n = 2, P(2) = 3u_1 - 2u_0 = 3.3 - 2.2 = 5 = 2^2 + 1 = 5\)

    Assume that \(P(m)\) and \(P(m + 1)\) is true i.e. \(P(m) = 2^m + 1\) and \(P(m + 1) = 2^{m + 1} + 1\)

    For \(n = m + 2.\)

    \(u_{m + 2} = 3u_{m + 1} - 2u_m\)

    \(= 3(2^{m + 1} + 1) - 2(2^m + 1)\)

    \(= 2.2^{m + 1} + 1 = 2^{m + 2} + 1\)

    Thus, \(P(n)\) is true for \(n = m + 2.\) Henece, we have proven the equation by mathematical induction.

  11. Let \(P(n) = a_n = 2^n - 1\)

    For \(n = 2, P(2) = 3a_1 - 2a_0 = 3.1 - 2.0 = 3 = 2^2 - 1 = 3\)

    Assume that \(P(m)\) and \(P(m + 1)\) is true i.e. \(P(m) = 2^m - 1\) and \(P(m + 1) = 2^{m + 1} - 1\)

    For \(n = m + 2,\)

    \(a_{m + 2} = 3a_{m + 1} - 2a_{m}\)

    \(= 3(2^{m + 1} - 3) -2(2^m + 1)\)

    \(= 2^{m + 2} - 1\)

    Thus, \(P(n)\) is true for \(n = m + 2.\) Henece, we have proven the equation by mathematical induction.

  12. Let \(P(n) = \cos n\theta, n > 2\)

    For \(n = 3, P(3) = 2A_2\cos\theta - A_1 = 2\cos 2\theta\cos\theta - \cos\theta\)

    \(= \cos\theta(2\cos^2\theta - 2\sin^2\theta - 1)\)

    \(= \cos\theta(\cos2\theta - 2\sin^2\theta)\)

    \(=\cos2\theta\cos2\theta - 2\sin2\theta\sin\theta = \cos3\theta\)

    Assume that \(P(m)\) and \(P(m + 1)\) is true i.e. \(P(m) = \cos m\theta\) and \(P(m + 1) = \cos(m + 1)\theta\)

    For \(n = m + 2,\)

    \(P(m + 2) = 2A_{m + 1}\cos\theta - A_m\)

    \(= 2\cos(m + 1)\theta\cos\theta - \cos m\theta\)

    We know that \(2\cos A\cos B = \cos(A + B) + \cos(A - B)\)

    \(P(m + 2) = \cos(m + 2)\theta + \cos m\theta - \cos m\theta\)

    \(P(m + 2) = \cos(m + 2)\theta\)

    Thus, \(P(n)\) is true for \(n = m + 2.\) Henece, we have proven the equation by mathematical induction.

  13. Let \(P(n) = (2\cos\theta - 1)(2\cos 2\theta - 1)\ldots(2\cos 2^{n - 1}\theta - 1) = \frac{2\cos 2^n\theta + 1}{2\cos\theta + 1}\)

    For \(n = 1, P(1) = 2\cos\theta - 1 = \frac{2\cos 2\theta + 1}{2\cos\theta + 1} = \frac{4\cos^2\theta - 1}{2\cos\theta + 1} = 2\cos\theta - 1\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = (2\cos\theta - 1)(2\cos 2\theta - 1)\ldots(2\cos 2^{m - 1}\theta - 1)(2\cos 2^m\theta - 1)\)

    Substituting the value of \(P(m)\)

    \(P(m + 1) = \frac{2\cos 2^m\theta + 1}{2\cos\theta + 1}(2\cos 2^m\theta - 1)\)

    \(= \frac{4\cos 2^m\theta)^2 - 1}{2\cos\theta + 1}\)

    \(= \frac{2\cos2^{m + 1}\theta + 1}{2\cos\theta + 1}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  14. Let \(P(n) = \tan^{-1}\frac{x}{1.2 + x^2} + \tan^{-1}\frac{x}{2.3 + x^2} + \ldots + \tan^{-1}\frac{x}{n(n + 1) + x^2} = \tan^{-1}x - \tan^{-1}\frac{x}{n + 1}\)

    We know that \(\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x - y}{1 + xy}\)

    For \(n = 1, P(1) = \tan^{-1}\frac{x}{1.2 + x^2} = \tan^{-1}x - \tan^{-1} \frac{x}{2} = \tan^{-1}\frac{x - \frac{x}{2}}{1 + \frac{x^2}{2}}\)

    \(= \tan^{-1}\frac{x}{2 + x^2}\)

    Thus, \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = \tan^{-1}\frac{x}{1.2 + x^2} + \tan^{-1}\frac{x}{2.3 + x^2} + \ldots + \tan^{-1}\frac{x}{m(m + 1) + x^2} + \tan^{-1}\frac{x}{(m +1)(m + 2) + x^2}\)

    Substituting the value of \(P(m)\)

    \(P(m + 1) = \tan^{-1}x - \tan^{-1}\frac{x}{m + 1} + \tan^{-1}\frac{x}{(m + 1)(m +2) + x^2}\)

    \(P(m + 1) = \tan^{-1}x - \tan^{-1}\frac{x}{m + 1} + \tan^{-1}\frac{x}{m + 1} - \tan^{-1}\frac{x}{m + 2}\)

    \(P(m + 1) = \tan^{-1}x - \tan^{-1}\frac{x}{m + 2}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  15. Let \(P(n) = 3 + 33 + \ldots + \frac{33\ldots3}{n~\text{digits}} = \frac{10^{n + 1} - 9n -10}{27}\)

    For \(n = 1, P(1) = 3 = \frac{10^2 - 9 -10}{27} = \frac{81}{27} = 3\)

    First term is \(3 = \frac{10^1 - 1}{3}\)

    Second terms is \(33 = \frac{10^2 - 1}{3}\)

    \(m{th}\) term is \(= \frac{10^m - 1}{3}\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = 3 + 33 + \ldots + \frac{33\ldots3}{m~\text{digits}} + \frac{33\ldots 3}{(m + 1)~\text{digits}}\)

    \(= \frac{10^{m + 1} - 9m -10}{27} + \frac{10^{m + 1} - 1}{3}\)

    \(= \frac{10^{m + 1} - 9m -10 + 9.10^{m + 1} - 9}{27}\)

    \(= \frac{10^{m + 1} - 9m -10 + (10 - 1).10^{m + 1} - 9}{27}\)

    \(= \frac{10^{m + 2} - 9(m + 1) - 10}{27}\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  16. Let \(P(n) = \int_{0}^{\pi}\frac{\sin(2n + 1)x}{\sin x}dx = \pi\)

    For \(n = 1, P(1) = \int_{0}^{\pi}\frac{\sin(2 + 1)x}{\sin x}dx\)

    \(= \int_{0}^{\pi} \frac{3\sin x - 4\sin^3x}{\sin x}dx\)

    \(= 3\int_{0}^{\pi}dx - 2\int_{0}^{\pi}dx + \int_{0}^{\pi}\cos 2xdx = \pi\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = \int_{0}^{\pi}\frac{\sin(2m + 3)x}{\sin x}dx\)

    \(P(m + 1) - P(m) = \int_{0}^{\pi}\frac{\sin(2m + 3)x - \sin(2m + 1)x}{\sin x}dx\)

    \(= \int_{0}^{\pi}\frac{2\cos(2m + 2)x\sin x}{\sin x}dx\)

    \(= \int_{0}^{\pi}2\cos(2m + 2)xdx = 0\)

    \(P(m + 1) = P(m) = \pi\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  17. Let \(P(n) = \int_{0}^{\pi}\frac{\sin^2 nx}{\sin^2x}dx = n\pi\)

    For \(n = 1, P(1) = \int_{0}^{\pi}\frac{\sin^2x}{\sin^2x}dx = \pi\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) For \(n= m + 1\)

    \(P(m + 1) = \int_{0}^{\pi}\frac{\sin^2(m + 1)x}{\sin^2x}dx\)

    \(P(m + 1) - P(m) = \int_{0}^{\pi}\frac{\sin^2(m + 1)x - \sin^2mx}{\sin^2x}dx\)

    \(= \frac{1}{2}\int_{0}^{\pi}\frac{1 - \cos(2m + 2)x -1 + \cos 2mx}{\sin^2x}dx\)

    \(= \frac{1}{2}\int_{0}^{\pi}\frac{\cos 2mx - \cos (2m + 2)x}{\sin^x}dx\)

    \(= \int_{0}^{\pi}\frac{2\sin(2m + 1)x\sin x}{\sin^2x}dx\)

    We have proven above to be \(\pi\) in previous example, therefore

    \(P(m + 1) = \pi + m\pi = (m + 1)\pi\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  18. This problem has been left as an exercise.

  19. Let \(P(n) = n(n + 1)(n + 5)\)

    For \(n = 1, p(1) = 1.2.6\) which is divisible by \(6\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m\) i.e. \(m(m + 1)(m + 5) = 6k\) where \(k\in N\)

    \(P(m + 1) = (m + 1)(m + 2)(m + 6) = (m + 1)(m + 2)(m + 6)\)

    \(P(m + 1) - P(m) = (m + 1)\left\{(m + 6)(m + 2) - m(m + 5)\right\}\)

    \(= (m + 1)(3m + 12) = 3(m + 1)(m + 4)\)

    Clearly, either \(m + 1\) or \(m + 4\) will be an even number making difference divisvible by \(6.\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  20. Let \(P(n) = n^3 + (n + 1)^3 + (n + 2)^3, n\in N\)

    For \(n = 1, P(1) = 1^3 + 2^3 + 3^3 = 36\) which is divisible by \(9\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = (m + 3)^3 - m^3 = 9m^2 + 27m + 27\) which is divisible by \(9\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  21. Let \(P(n) = n(n^2 + 20), n=2m, m\in N\)

    For \(n = 2, P(2) = 2(2^2 + 20)\) which is divisible by \(48\)

    \(P(n)\) is true for \(n = 2.\) Assume that it is true for \(n = 2m.\) For \(n = 2m + 2\)

    \(P(2m + 2) = (2m + 2)\left[(2m + 2)^2 + 20\right]\)

    \(P(2m + 2) - P(2m) = (2m + 2)^3 + (2m + 2)20 - (2m)^3 - 40m\)

    \(= 24m(m + 1)+ 48\)

    Clearly, either \(m\) or \(m + 1\) will be even making above expression divisible by \(48\)

    Thus, \(P(n)\) is true for \(n = 2m + 2.\) Henece, we have proven the equation by mathematical induction.

  22. Let \(P(n) = 4^n - 3n - 1, n\in N\)

    For \(n = 1, P(1) = 4^1 - 3.1 -1 = 0\) which is divisible by \(9\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) = 4^{m + 1} -3(m + 1) - 1\)

    \(P(m + 1) - P(m) = 4^{m + 1} - 3(m + 1) - 1 - 4^m + 3m + 1\)

    \(= 3.4^m - 3 = 3(4^m - 1^m) = 3.(4 - 1)(4^{m - 1} + \ldots)\)

    Clearly, the above expression is divisible by \(9\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  23. Let \(P(n) = 3^{2n} - 1, n\in N\)

    For \(n = 1, P(1) = 3^2 - 1 = 8\) which is divisible by \(8\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = 3^{2m + 2} - 1 - 3^2m + 1\)

    \(= 3^2m(3^2 - 1)\) which is divisible by \(8\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  24. Let \(P(n) = 5.2^{3n - 2} + 3^{3n - 1}, n\in N\)

    For \(n = 1, P(1) = 5.2^{3.1 - 2} + 3^{3 - 1} = 19\) which is divisible by \(19\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = 5.2^{3m + 1} + 3^{3m + 2} - 5.2^{3m - 2} - 3^{3m - 1}\)

    \(= 5.2^{3m - 2}.7 + 3^{3m - 1}.26 = 7P(m) + 10.3^{3m - 1}\)

    which is divisible by \(19\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  25. Let \(P(n) = 7^{2n} + 2^{3n - 3}.3^{n - 1}, n\in N\)

    For \(n = 1, P(1) = 7^{2.1} + 2^{3.1 - 3}.3^{1 - 1} = 49 + 1 = 50\) which is divisible by \(25\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = 7^{2m + 2} + 2^{3m}.3^{m} - 7^{2m} - 2^{3m - 3}.3^{m - 1}\)

    \(= 48.7^{2m} + 2^{3m - 3}.3^{m - 1}(8.3 - 1)\)

    \(= 23.P(m) + 25.7^{2m}\) which is divisible by \(25\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  26. Let \(P(n) = 10^n+ 3.4^{n + 2} + 5, n\in N\)

    For \(n = 1, P(1) = 10^1 + 3.4^{1 + 2} + 5 = 207\) which is divisibleby \(9\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = 10^{m + 1} + 3.4^{m + 3} + 5 - 10^{m} - 3.4^{m + 2} - 5\)

    \(= 9.10^m + 9.4^{m + 2}\) which is divisible by \(9\)

    Thus, \(P(n)\) is true for \(n = m + 1.\) Henece, we have proven the equation by mathematical induction.

  27. Let \(P(n) = 3^{4n + 2} + 5^{2n + 1}, n\in N\)

    For \(n = 1, P(1) = 3^{6} + 5^3 = 854\) which is divisible by \(14\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = 3^{4m + 6} + 5^{2m + 3} - 3^{4m + 2} - 5^{2m + 1}\)

    \(= 80.3^{4m + 2} + 24.5^{2m + 1} = 24P(m) + 56.3^{4m + 2}\) which is divisible by \(14\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

  28. Let \(P(n) = 3^{2n + 2} - 8n - 9, n\in N\)

    For \(n = 1, P(1) = 3^{4} - 8 - 9 = 64\) which is divisible by \(64\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = 3^{2m + 4} - 8(m + 1) - 9 - 3^{2m + 2} + 8m + 9\)

    \(= 8.3^{2m + 2} - 8 = 64.3^{2m}\) which is divisible by \(64\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

  29. Let \(P(n) = n^7 - n, n\in N\)

    For \(n = 1, P(1) = 1 - 1 = 0\) which is a multiple of \(7\)

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

    \(P(m + 1) - P(m) = (m + 1)^7 - m - 1 - m^7 + m\)

    \(= {}^7C_1m^6 + {}^7C_2m^5 + \ldots + {}^7C_6m\) which is a multiple of \(7\) [This uses binomial expansion which is next chapter.]

    \(P(n)\) is true for \(n = 1.\) Assume that it is true for \(n = m.\) For \(n = m + 1\)

Rest of the problems are left as exercises.