The sum of the first $20$ terms of the series $5+11+19+29+41+\ldots$ is :

$$ \begin{aligned} & \mathrm{S}_n=5+11+19+29+41+\ldots .+\mathrm{T}_n \\\\ & \mathrm{S}_n=~~~~~~~~ 5+11+19+29+\ldots .+\mathrm{T}_{n-1}+\mathrm{T}_n \\\\ & \hline 0=5+6+8+10+12+\ldots . . \mathrm{T}_n \end{aligned} $$ <br/><br/>$$ \begin{array}{rlrl} &0 =5+\frac{n-1}{2}[2 \times 6+(n-2)(2)]-T_n \\\\ &\Rightarrow T_n =5+(n-1)(n+4) \\\\ &\Rightarrow T_n =5+n^2+3 n-4 \\\\ &\Rightarrow T_n =n^2+3 n+1 \\\\ &\Sigma T_n =\Sigma n^2+3 \Sigma n+\Sigma 1 \end{array} $$ <br/><br/>$\Rightarrow \quad S_n=\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}+n$ <br/><br/>When, $n=20$ <br/><br/>Then, <br/><br/>$$ \begin{aligned} S_{20} & =\frac{20 \times 21 \times 41}{6}+\frac{3 \times 20 \times 21}{2}+20 \\\\ & =2870+630+20=3520 \end{aligned} $$