Question

If the sum of n terms of series is $1^2+2\times 2^2+3^2+2\times 4^2+5^2+2\times 6^2+....is\frac{n(n+1^2)}{2}$ when n is even, then when n is odd, then s

• A. $\frac{n(n+1^2)}{4}$
• B. $\frac{n^2(n+1)}{2}$
• C. $\frac{3n(n+1)}{2}$
• D. $[\frac{n(n+1)}{2}{]}^2$

Answer 1

The correct option is B $\frac{n^2(n+1)}{2}$

$1^2+2\times 2^2+3^2+2\times 4^2+2\times 4^2+5^2+2\times 6^2+------\frac{m(n+1)}{2}$

$1+2\times 2^2+3^2+{2.4}^2+-------2\left(2m{)}^{+}\right(2m+1{)}^2$

$\sum (2m+1{)}^2+4[1^2+2^2+3^2------m^2]$

$=(2m+1)(2m+2)(4m+2+1)/64m(m+1)(2m+1)/6$

$=(2m+1)(m+1)/6\left[2\right(4m+3)+4m]$

$=(2m+1)(2m+2)(6m+3)/6$

$=(2m+1{)}^2(2m+2)/2=n^2(n+1)/2$