Question

Show by the Principle of Mathematical induction that the sum $S_n$ of the n 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 $S_n=\{\begin{array}{cc}\frac{n(n+1{)}^2}{2},\text{if n is even}& \\ \frac{n^2(n+1)}{2},\text{if n is odd}& \end{array}$

Answer 1

$S_n=1^2+2\times 2^2+3^2+2\times 4^2+.......$ Using induction we first show this is true for n = 2,

We get $S_2=1^2+2\times 2^2=1+8=9$

From RHS, we have if n is even $S_n=\frac{n(n+1{)}^2}{2}$

$S_2=\frac{2\times 9}{2}=9$

Now using induction we first show this is true also for n = 3, we get $S_3=1+8+9=18$

From RHS, we have if n is odd $S_n=\frac{n^2(n+1)}{2}$

$S_3=\frac{9\times 4}{2}=18$

Let assume above is true for n = k we get

k is even, $S_k=1^2+2\times 2^2+3^2+2\times 4^2+....+2\times k^2$ ........(i)

k is odd, $S_k=1^2+2\times 2^2+3^2+2\times 4^2+.....k^2$ ........(ii)

Now lets prove join = k + 1

If k is even, k + 1 is odd we get

$S_{k+1}=1^2+2\times 2^2+3^2+2\times 4^2+......\times 2\times k^2=\frac{k(k+1{)}^2}{2}$

Substitute this in 3, we get

$S_{k+1}=\frac{k(k+1{)}^2}{2}+(k+1{)}^2$

$=\frac{(k+1{)}^2(k+2)}{2}$

$=RHS$ (when 'k + 1' is odd)

Hence proved.