Question

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

• 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
$S_n=5+11+19+29+...+a_{n-1}+a_n...\left(i\right)$
or $S_n=5+11+19+...+a_{n-2}+a_{n-1}+a_n...\left(ii\right)$
Subtracting $\left(i\right)$ and $\left(ii\right)$, we get
$0=5+[6+8+10+12+...\left(n-1\right)$ terms$]-a_n$
or $a_n=5+\frac{\left(n-1\right)\left[12+\left(n-2\right)\times 2\right]}{2}$
$=5+\left(n-1\right)\left(n+4\right)=n^2+3n+1$
Hence $S_n=\sum _{k=1}^n{a}_k$
$=\sum _{k=1}^n\left(k^2+3k+1\right)$
$=\sum _{k=1}^n{k}^2+3\sum _{k=1}^{n}k+n$
$=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{3n\left(n+1\right)}{2}+n$
$=\frac{n\left(n+2\right)\left(n+4\right)}{3}$.