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

Answer 1

Answer: (A)

Given series is $ {{1.3}^{2}}+{{2.5}^{2}}+{{3.7}^{2}}+...\text{ }\infty $ This is an arithmetic-geometric series whose nth term is equal to $ {{T}_{n}}=n{{(2n+1)}^{2}}+4{{n}^{3}}+4{{n}^{2}}+n $ $ \therefore $ $ {{S}_{n}}=\sum\limits_{1}^{n}{{{T}_{n}}}=\sum\limits_{1}^{n}{(4{{n}^{3}}+4{{n}^{2}}+n)} $ $ =4\sum\limits_{1}^{n}{{{n}^{3}}}+4\sum\limits_{1}^{n}{{{n}^{2}}}+\sum\limits_{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)(6{{n}^{2}}+14n+7) $