Question

There are n A.P.'s whose common differences are $1,2,3,$_________n respectively, the first term of each being unity. Prove that sum of their nth terms is $\frac{1}{2}n(n^2+1)$.

Answer 1

Here $a=1$ for all, and $d=1,2,3,...,n$ respectively for the n A.P.'s
We have to find sum of their nth terms.
$\sum T_n=[1+(n-1\left).1\right]+[1+(n-1\left).2\right]+...+[1+(n-1).n]$
$=[1+1+1+...nterms]+(n-1)[1+2+3+...nterms]$
$=n+(n-1)\frac{n}{2}[1+n]=\frac{n}{2}(2+n^2-1)$
$=\frac{n}{2}[n^2+1]$.