Question

The sum of the series $3+8+16+27+\mathrm{41.......}$ 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,
$0=3+5+8+11+........+\left(t_n-t_{n-1}\right)-t_n$
$\Rightarrow t_n=1+\left(2+5+8+11+.........nterms\right)$
$=1+\frac{n}{2}\left(4+\left(n-1\right)3\right)=1+\frac{n}{2}\left(3n+1\right)$
$S_n=\Sigma t_n=\Sigma 1+\frac{3}{2}\Sigma n^2+\frac{1}{2}\Sigma n$
$=\frac{3}{2}\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{1}{2}\frac{n\left(n+1\right)}{2}+n$
for $n=20,S=\frac{20\times 21\times 41}{4}+\frac{20\times 21}{4}+20$
$=\frac{20\times 21}{4}\left(41+1\right)+20=4430$