Question

Find the sum of the series $3+7+14+24+37+......$up to $10$ terms.

• A. $570$
• B. $750$
• C. $705$
• D. $507$

Answer 1

Answer: (A)

$570$

$3+7+14+24+\mathrm{37.....}$ upto $10$ terms

Let general terms of this series be

$t_n=a{n}^2+bn+c$

$a+b+c=3---\left(1\right)$

$4a+2b+c=7---\left(2\right)$

$9a+3b+c=14---\left(3\right)$

on solving, $a=\frac{3}{2}b=\frac{-1}{2}c=2$

$\sum _{n=1}^{10}\frac{3}{2}{n}^2\frac{-1}{2}n+2$

$\frac{3}{2}\frac{n(n+1)(2n+1)}{6}\frac{-1}{2}\frac{n(n+1)}{2}+2n$

$=\frac{3\times 10\times 11\times 21}{2\times 6}-\frac{55}{2}+20$

$=\frac{1100}{2}+20$

$=570$

$A$ is correct.