The sum of the first n terms of the series `1^2+2xx2^2+3^2+2xx 4^2+5^2+2xx6^2…..is (n(n+1)^2)/(2)` when n is even .Then find the sum when n is odd.

The sum of the first n terms of the series 1^2+ 2.2^2+3^2 +2.4^2+.... is (n(n+1)^2)/2 when n is even. Then the sum if n is odd , is

Find the sum of n terms of the series 1. 2^2+2. 3^2+3. 4^2+

Assertion: If n is odd then the sum of n terms of the series 1^2+2xx2^2+3^2+2xx4^2+5^2+2xx6^2+7^2+…is (n^2(n+1))/2. If n is even then the sum of n terms of the series. 1^2+2xx2^2+3^2+2xx4^2+5^2+2xx6^2+…..is (n(n+1)^2)/2 (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

Find the sum to n terms of the series 1^2+2^2+3^2-4^2+5^2-6^2+. . . .

Show by the Principle of Mathematical induction that the sum S_n , of the nterms of the series 1^2 + 2xx 2^2 + 3^2 + 2xx 4^2+5^2 +2 xx 6^2 +7^2+..... is given by S_n={(n(n+1)^2)/2 , if n is even , then (n^2(n+1))/2 , if n is odd

Sum of n terms the series : 1^2-2^2+3^2-4^2+5^2-6^2+

Sum of n terms the series : 1^2-2^2+3^2-4^2+5^2-6^2+....

Find the sum of first n terms of the series 1^3+3xx2^2+3^3+3xx4^2+5^3+3xx6^2+ w h e n n is even n is odd

Find the sum to n terms of the series 1xx2+2xx3+ 3xx4+ 4xx5+......

Find the sum to n terms of the series 1xx2+2xx3+ 3xx4+ 4xx5+......