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

Answer

`[(n^(2) (n+1))/(2)]`

Explanation

Here, `1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) + 5^(2) +`... upto n terms
`= (n (n +1)^(2))/(2)` [where n is even] ...(i)
When n is odd, `1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) = 5^(2) ... + n^(2)`
`= {1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) + ...+ 2 (n -1)^(2)} + n^(2)`
`= {((n -1) (n)^(2))/(2)} + n^(2)` [from Eq. (i)]
`= n^(2) ((n-1)/(2) + 1) = n^(2) ((n+1))/(2)`
`:. 1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) +`.... upto n terms, when n is odd
`= (n^(2) (n+1))/(2)`