Question

The sum of the series $3+8+16+27+41.......$ upto $20$ terms is equal to

• A. $4230$
• B. $4430$
• C. $4330$
• D. $4500$

Answer 1

Answer: (B)

$S_{n}=3+8+16+27+........+t_{n}$
$S_{n}=3+8+16+........+t_{n - 1}+t_{n}$
Subtracting, we get,
$0=3+5+8+11+........+\left(t_{n} - t_{n - 1}\right)-t_{n}$
$\Rightarrow t_{n}=1+\left(2 + 5 + 8 + 11 + . . . . . . . . . n t e r m s\right)$
$=1+\frac{n}{2}\left(4 + \left(n - 1\right) 3\right)=1+\frac{n}{2}\left(3 n + 1\right)$
$S_{n}=\Sigma t_{n}=\Sigma 1+\frac{3}{2}\Sigma n^{2}+\frac{1}{2}\Sigma n$
$=\frac{3}{2}\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}+\frac{1}{2}\frac{n \left(n + 1\right)}{2}+n$
for $n=20,S=\frac{20 \times 21 \times 41}{4}+\frac{20 \times 21}{4}+20$
$=\frac{20 \times 21}{4}\left(41 + 1\right)+20=4430$