The sum of the first `1^2+2.2^2+3^2+2.4^2+5^2+2.6^2.....`is `(n^2(n+1))/2` when n is odd.Then the sum of first 20 terms is

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

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.

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

The sum of the series 1^2+3^2+5^2+ ........ n

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

Write the sum of the series 1^2-2^2+3^2-4^2+5^2-6^2...........+(2n-1)^2-(2n)^2dot

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

The sum of the series 1^(2)+1+2^(2)+2+3^(2)+3+ . . . . .. +n^(n)+n , is

The sum of the series 1^(2)+1+2^(2)+2+3^(2)+3+ . . . . .. +n^(2)+n , is

The sum of the squares of the first n positive integers 1^2 + 2^2 + 3^2 + ………+ n^2 is (n(n +1)(2n+1))/(6) . What is the sum of the squares of the first 9 positive integers?