The sum of the first 20 terms of the series [5+11+19+29+41+] is :

Question

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

Accepted Answer

$$
\begin{aligned}
& \mathrm{S}_n=5+11+19+29+41+\ldots .+\mathrm{T}_n \\
& \mathrm{S}_n=~~~~~~~~ 5+11+19+29+\ldots .+\mathrm{T}_{n-1}+\mathrm{T}_n \\
&  0=5+6+8+10+12+\ldots . . \mathrm{T}_n
\end{aligned}
$$

$$
\[\begin{gathered} 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{gathered}$$
\]

$$
\Rightarrow \quad S_n=\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}+n
$$

When, $n=20$

Then,

$$
\begin{aligned}
S_{20} & =\frac{20 \times 21 \times 41}{6}+\frac{3 \times 20 \times 21}{2}+20 \\
& =2870+630+20=3520
\end{aligned}
$$

Solutions · 1