Question

Show by the Principle of Mathematical induction that the sum S_n of then terms of the series  \[1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + . . .\] is given by \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]

 

Answer 1

\[ \text{ Let } P\left( n \right): S_n = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ when n is even } }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ when n is odd} }\]
\[\text{ Step I: For } n = 1 i . e . P\left( 1 \right): \]
\[LHS = S_1 = 1^2 = 1\]
\[RHS = S_1 = \frac{1^2 \left( 1 + 1 \right)}{2} = 1\]
\[\text{ As, LHS = RHS } \]
\[\text{ So, it is true for n }  = 1 . \]
\[ \text{ Step II: For n }  = k, \]
\[\text{ Let } P\left( k \right): S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . = \binom{\frac{k \left( k + 1 \right)^2}{2}, \text{ when k is even} }{\frac{k^2 \left( k + 1 \right)}{2}, \text{ when k is odd }}, \text{ be true for some natural numbers } . \]
\[\text{ Step III: For }  n = k + 1, \]
\[\text{ Case 1: When k is odd, then } \left( k + 1 \right) \text{ is even .}  \]
\[P\left( k + 1 \right): \]
\[LHS = S_{k + 1} = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . + k^2 + 2 \times \left( k + 1 \right)^2 \]
\[ = \frac{k^2 \left( k + 1 \right)}{2} + 2 \times \left( k + 1 \right)^2 \left( \text{ Using step }  II \right)\]
\[ = \frac{k^2 \left( k + 1 \right) + 4 \left( k + 1 \right)^2}{2}\]
\[ = \frac{\left( k + 1 \right)\left( k^2 + 4k + 4 \right)}{2}\]
\[ = \frac{\left( k + 1 \right) \left( k + 2 \right)^2}{2}\]
\[RHS = \frac{\left( k + 1 \right) \left( k + 1 + 1 \right)^2}{2}\]
\[ = \frac{\left( k + 1 \right) \left( k + 2 \right)^2}{2}\]
\[\text{ As, LHS = RHS } \]
\[\text{ So, it is true for n = k + 1 when k is odd .}  \]
\[\text{ Case 2: When k is even, then }  \left( k + 1 \right) \text{ is odd .}  \]
\[P\left( k + 1 \right): \]
\[RHS = S_{k + 1} = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . + 2 \times k^2 + \left( k + 1 \right)^2 \]
\[ = \frac{k \left( k + 1 \right)^2}{2} + \left( k + 1 \right)^2 \left( \text{ Using step } II \right)\]
\[ = \frac{k \left( k + 1 \right)^2 + 2 \left( k + 1 \right)^2}{2}\]
\[ = \frac{\left( k + 1 \right)^2 \left( k + 2 \right)}{2}\]
\[ = \frac{\left( k + 1 \right)^2 \left( k + 2 \right)}{2}\]
\[RHS = \frac{\left( k + 1 \right)^2 \left( k + 1 + 1 \right)}{2}\]
\[ = \frac{\left( k + 1 \right)^2 \left( k + 2 \right)}{2}\]
\[As, LHS = RHS\]
\[\text{ So, it is true for n = k + 1 when k is even }  . \]
\[\text{ Hence, } P\left( n \right) \text{ is true for all natural numbers .}  \]