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

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

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.

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 of n terms of the series 1^3+3.2^2+3^3+3.4^2+5^3+3.6^2+....... when (i)n is odd (ii)n is even

Find the sum of n terms of the series 1^2 + 4^2+ 7^2 + .........

Find the sum of n terms of the series 1^2 + 4^2+ 7^2 + .........

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+... If B-2A=100lambda then lambda is equal to (1) 232 (2) 248 (3) 464 (4)496

Find the sum to n terms of the series : 1^2+(1^2+2^2)+(1^2+2^2+3^2)+.......

Sum up to n terms the series 1.2^(2)+2.3^(2)+ 3.4^(2)+...

Statement 1 The sum of first n terms of the series 1^(2)-2^(2)+3^(2)-4^(2)_5^(2)-"……" can be =+-(n(n+1))/(2) . Statement 2 Sum of first n narural numbers is (n(n+1))/(2)