Question

The sum to n terms of the infinite series ${1.3}^2+{2.5}^2+{3.7}^2+...\infty$ is

• A. $\frac{n}{6}(n+1)(6n^2+14n+7)$
• B. $\frac{n}{6}(n+1)(2n+1)(3n+1)$
• C. $4n^3+4n^2+n$
• D. None of the above

Answer 1

Answer: (A)

Given series is ${1.3}^2+{2.5}^2+{3.7}^2+...\infty$ This is an arithmetic-geometric series whose nth term is equal to $T_n=n{(2n+1)}^2+4n^3+4n^2+n$ $\therefore$ $S_n=\sum _1^n{T}_n=\sum _1^n(4n^3+4n^2+n)$ $=4\sum _1^n{n}^3+4\sum _1^n{n}^2+\sum _1^{n}n$ $=4{\left(\frac{n}{2}(n+1)\right)}^2+\frac{4}{6}n(n+1)(2n+1)+\frac{n}{2}(n+1)$ $=n(n+1)\left[n^2+n+\frac{4}{6}(2n+1)+\frac{1}{2}\right]$ $=\frac{n}{6}(n+1)(6n^2+14n+7)$