The sum of the first n terms of the series 1^2+22^2+3^2+24^2+5^2+26^2+ is n(n+1)^22 when n…

Question

The sum of the first $n$ terms of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+\cdots$ is $\dfrac{n(n+1)^2}{2}$ when $n$ is even. When $n$ is odd, the sum is:

Accepted Answer

If $n$ is odd, the required sum is

$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$

$$ = {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$$

[  As $$\left( {n - 1} \right)$$ is even

$$\therefore$$ using given formula for the sum of $$\left( {n - 1} \right)$$ terms.]

$$ = \left( {{{n - 1} \over 2} + 1} \right){n^2} = {{{n^2}\left( {n + 1} \right)} \over 2}$$

Solutions · 1