Question

Find the sum of terms of the sequence till n terms : $6+9+16+27+42+....+n$ $terms$.

Answer 1

$S=6+9+16+27+42+.......n$ terms

$S=6+9+16+27+.............$

$O=6+(3+7+11+15+...n+terms)$- $T_n$

$T_n=6+\frac{n-1}{2}(2.3+(n-2\left)4\right)$

$T_n=6+\frac{(n-1)(4n-2}{2}$

$T_n=6+(n-1)(2n-1)$

$T_n=2n^2-3n+7$

$S\sum _{r=1}^n{T}_r=2\sum _{r=1}^n{r}^2+7\sum _{r=1}^{n}1-3\sum _{r=1}^{n}r$

$=\frac{2\left(n\right)(n+1)(2n+1)}{6}+7n\frac{3n(n+1)}{2}$

$=\frac{2n(n+1)(2n+1)+42n-9n(n+1)}{6}$

$=\frac{4n^3+2n^2+4n^2+2n+42n-9n^2-9n}{6}$

$=\frac{4n^2-3n^2+35n}{6}$