Question

Sum of the series $4+6+9+13+18+......n$ terms , is

Answer 1

$a_1=4=3+1$

$a_2=6=a_1+2$

$a_3=9=a_2+3$

$a_4=13=a_3+4$

$a_{n+1}=a_{n-2}+(n-1)$

$a_n=a_n+n$

on adding all terms

$(a_1+a_2+a_3.....a_n)=(a_1+a_2+a_3+....a_{n+1}+3+(1+2+3+.....n)$

$\Rightarrow a_n=3+\sum n$

$\Rightarrow a_n=3+\frac{n(n+1)}{2}$

$4+6+9+13+.....nterms\Rightarrow \sum a_n$

$\Rightarrow {\sum }_{n=1}^n(3+\frac{n(n+1)}{2})$

$\Rightarrow 3n+{\sum }_{n=7}^n\frac{n^2}{2}+{\sum }_{n=1}^n\frac{n}{2}$

$\Rightarrow 3n+\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{1}{2}\frac{n(n+1)}{2}$

$\Rightarrow \frac{36n+n(n+1)(2n+1)+3n(n+1)}{12}$

$4+6+9+13+....nterms\Rightarrow \frac{n}{12}[36+3n+3+2n^2+3n+1]$

$4+6+9+\mathrm{13....}nterms\Rightarrow \frac{n}{6}(n^2+3n+20)$