The sum of `n` terms of the series `1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+....` is`(n(n+1)^2)/2` when n is even . when n is odd , the sum is

Options

A. `(3n(n+1))/2`
B. `(n^(2)(n+1))/2`
C. ` (n(n+1)^(2))/4`
D. ` [ (n(n+1))/2]^(2)`

Answer

b

Explanation

The sum of n terms of given series = ` (n(n+1)^(2))/2`
if n is even .
Let n be odd i.e n =` 2m+1`
Then, ` S_(2m+1)=S_(2m)+(2m+1)` th them
` = ((n-1)n^(2))/2 + n " th "term = ((n-1)n^(2))/2 + n^(2`
` [ :' ` n is odd = `2m+1` ]
` = n^(2) [ (n-1+2)/2] = ((n+1)n^(2))/2`