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

Let the given series be $S_n$, then we can write as

$$\begin{gathered} S_n=5+11+19+29+41+............+T_n \\ {S}_n= 5+11+19+29+41+......+T_{n-1}+T_n \end{gathered}$$

Subtracting above equations, we get

$0=5+6+8+10+....-T_n$

$\Rightarrow T_n=5+\left[\frac{\left(n-1\right)}{2}\left(12+\left(n-2\right)2\right)\right]$

$\Rightarrow T_n=5+\left(n-1\right)\left(n+4\right)$

$\Rightarrow T_n=n^2+3n+1$

So,

$S_{20}=\sum _{n=1}^{20}{T}_n$

$\Rightarrow S_{20}=\sum _{n=1}^{20}\left(n^2+3n+1\right)$

$\Rightarrow S_{20}=\frac{20\times 21\times 41}{6}+\frac{3\times 20\times 21}{2}+20$

$\Rightarrow S_{20}=3520$