Question

Find the sum to $n$ terms of the series:
$5 + 11 + 19 + 29 + 41 ... $

• A. $\frac{n\left(n+2\right)\left(n+4\right)}{3}$
• B. $\frac{n\left(n+1\right)\left(n+2\right)}{3}$
• C. $\frac{n\left(n+2\right)\left(n+3\right)}{3}$
• D. None of these

Answer 1

Answer: (A)

Let us write
$S_{n} = 5+ 11 + 19 + 29 + ... +a_{n-1}+ a_{n} \quad...\left(i\right)$
or $S_{n}= 5 +11 + 19 + ...+ a_{n-2}+a_{n-1} + a_{n} \quad...\left(ii\right)$
Subtracting $\left(i\right)$ and $\left(ii\right)$, we get
$0 = 5 +[6 + 8+ 10+12 + ...\left(n- 1\right)$ terms$] - a_{n}$
or $a_{n}=5+\frac{\left(n-1\right)\left[12+\left(n-2\right)\times2\right]}{2}$
$ = 5+ \left(n-1\right)\left(n+4\right) = n^{2} +3n +1 $
Hence $S_{n}= \sum\limits_{k=1}^{n}a_{k} $
$= \sum\limits_{k=1}^{n}\left(k^{2} +3k +1\right) $
$= \sum\limits_{k=1}^{n} k^{2} +3\sum\limits ^{n}_{k=1} k+n $
$ = \frac{n\left(n+1\right)\left(2n+1\right)}{6} + \frac{3n\left(n+1\right)}{2} +n$
$ = \frac{n\left(n+2\right)\left(n+4\right)}{3}$.