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.When n is odd the sum is -

• A. $\frac{3n(n+1)}{2}$
• B. $\frac{8n^3+6n^2+n}{3}$
• C. $\frac{n(n+1{)}^2}{4}$
• D. ${\left[\frac{n(n+1)}{4}\right]}^2$

Answer 1

Answer: (B)

$\frac{8n^3+6n^2+n}{3}$

$1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+.......(1^2+3^2+5^2+.....n^2)+2(2^2+4^2+6^2+......)=(1^2+3^2+5^2+.....n^2)+2.2(1^2+2^2+3^2+......)=(1^2+3^2+5^2+.....n^2)+2^2(1^2+2^2+3^2+......)=\frac{n(4n^2-1)}{3}+4\left[\frac{n(n+1)(2n+1)}{6}\right]=\frac{n(4n^2-1)}{3}+\frac{2(n^2+n)(2n+1)}{3}=\frac{4n^3-n+(4n+2)(n^2+n)}{3}=\frac{4n^3-n+4n^3+4n^2+2n^2+2n}{3}=\frac{8n^3+6n^2+n}{3}$