Question

The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + ... is

Answer 1

The answer is $3520$

$T_{n}=an^{2}+bn+c$
$T_{1}=a+b+c=5$
$T_{2}=4a+2b+c=11$
$T_{3}=9a+3b+c=19$
$T_{n}=n^{2}+3n+1$
$S_{20}=\sum_{n=1}^{20}T_{n}=\sum_{n=1}^{20}(n^{2}+3n+1)$
$S_{20}=\frac{20\times 21\times 41}{6}+3\times\frac{20\times 21}{2}+20$
$S_{20}=2870+630+20$
$S_{20}=3520$