Question

the sum of the first n terms of the series $1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2....is\frac{n(n+1{)}^2}{2}$ when n is even.wheen n is odd the sum is

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

Answer 1

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

$\text{When}n\text{is even}{S}_n=\frac{\left(n\right)(n+1{)}^2}{2}$

$S_n=1^2+2\cdot 2^2+3^2+-\cdots -(n-1{)}^2+2n^2$

When $n$ is odd,

$S_n=1^2+2\cdot 2^2+3^2+\cdots -2(n-1{)}^2+n^2$

$S_{n-1}=(\frac{(n-1)(n-1+1{)}^2}{2}=\frac{n^2(n-1)}{2})$

$S_n=\frac{n^2(n-1)}{2}+n^2=\frac{n^3-n^2+2n^2}{2}$

$S_n=\frac{n^2(n+1)}{2}$