Question

Find the $n^{th}$ term and sum to $n$ terms of the following series:
$1+9+24+46+75+$_____

Answer 1

$a_1=1$

$a_2=1+8=9$

$a_3=9+8+7=24$

$a_4=24+8+7+7=46$

So, then $k^{th}$ term will be:

$a_k=1+8(k-1)+7\frac{(k-1)(k-2)}{2}=\frac{7}{2}{k}^2-\frac{5}{2}k$

Sum of first $n$ terms

$=\sum _{k=1}^n(\frac{7}{2}{k}^2-\frac{5}{2}k)$

$=\frac{7}{2}\sum _{k=1}^n{k}^2-\frac{5}{2}\sum _{k=1}^{n}k$

$=\frac{7}{2}.\frac{1}{6}n(n+1)(2n+1)-\frac{5}{2}.\frac{1}{2}n(n+1)$

$=\frac{1}{12}n(n+1).\left(7\right(2n+1)-15)$

$=\frac{1}{12}n(n+1)(14n-8)$

$=\frac{1}{6}n(n+1)(7n-4)$