Question

Find the sum to $n$ terms of the series: $5+11+19+29+41\dots$

Answer 1

$S_n=5+11+19+29+41+...............=(1+2^2)+(2+3^2)+(3+4^2)+(4+5^2)+(5+6^2)+.......a_n=n+(n+1{)}^2=n+n^2+2n+1=n^2+3n+1\therefore S_n=\sum n^2+3n+1=(\frac{n(n+1)(2n+1)}{6})+(\frac{3n(n+1)}{2})+n=n\left[\right(\frac{2n^2+n+2n+1+9n+9+6}{6}\left)\right]=n\times (\frac{2n^2+12n+16}{6})=n\times (\frac{n^2+6n+8}{3})=(\frac{n(n+2)(n+4)}{3})$