For n ∈ N, the sum of the series 2.3 + 3.5 + 4.7 + ..... + (n + 1) (

1.	For n ∈ N, the sum of the series 2.3 + 3.5 + 4.7 + ..... + (n + 1) (2n + 1) is equal to(a) \(\frac{n}{3}(2n^2+3n+1)\) (b) \(\frac{1}{6}n(n^2+n-1)\)(c) \(\frac{n}{3}(3n^2+5n+11)\)(d) \(\frac{n}{6}(4n^2+15n+17)\)
Answer» (d) \(\frac{n}{6}(4n^2+15n+17)\) Let S = 2·3 + 3·5 + 4·7 + ..... + (n + 1) (2n + 1) \(\displaystyle\sum_{k=1}^{n}(k+1)\)(2k + 1) = \(\displaystyle\sum_{k=1}^{n}(2k^2+3k+1)\) = \(2\displaystyle\sum_{k=1}^{n}k^2\) + \(2\displaystyle\sum_{k=1}^{n}k+n\) = 2 x \(\frac{n(n+1)(2n+1)}{6}\) + 3 x \(\frac{n(n+1)}{2}\) + n = \(\frac{n}{6}\)[2(n + 1) (2n + 1) + 9 (n + 1) + 6] = \(\frac{n}{6}\) [4n² + 6n + 2 + 9 + 6] = \(\frac{n}{6}(4n^2+15n+17)\)

For n ∈ N, the sum of the series 2.3 + 3.5 + 4.7 + ..... + (n + 1) (2n + 1) is equal to(a) \(\frac{n}{3}(2n^2+3n+1)\) (b) \(\frac{1}{6}n(n^2+n-1)\)(c) \(\frac{n}{3}(3n^2+5n+11)\)(d) \(\frac{n}{6}(4n^2+15n+17)\)

Answer»

(d) \(\frac{n}{6}(4n^2+15n+17)\)

Let

S = 2·3 + 3·5 + 4·7 + ..... + (n + 1) (2n + 1)

\(\displaystyle\sum_{k=1}^{n}(k+1)\)(2k + 1) = \(\displaystyle\sum_{k=1}^{n}(2k^2+3k+1)\)

= \(2\displaystyle\sum_{k=1}^{n}k^2\) + \(2\displaystyle\sum_{k=1}^{n}k+n\)

= 2 x \(\frac{n(n+1)(2n+1)}{6}\) + 3 x \(\frac{n(n+1)}{2}\) + n

= \(\frac{n}{6}\)[2(n + 1) (2n + 1) + 9 (n + 1) + 6]

= \(\frac{n}{6}\) [4n² + 6n + 2 + 9 + 6]

= \(\frac{n}{6}(4n^2+15n+17)\)