The sum of the first n terms of the series `1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+...` is

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

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

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 the first n terms of the series 1^3+3.2^2+3^3+3.4^2+5^2+3.6^2+.... when n is odd

Find the sum of the first n terms of the series 1^3+3.2^2+3^3+3.4^2+5^2+3.6^2+.... when n is even

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=100 lambda then lambda is equal to (1)232(2)248 (3) 464 (4) 496

The sum of first 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 :

The sum of first 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 odd the sum is