Question

sum the series 1+(1+2)+(1+2+3)+...n terms.

Answer 1

n th term is =n(n+1)/2=n²+n/2=n²/2+n/2

submission 
1/2εn²+1/2∈n=n(n+1)(2n+1)/12+n(n+1)/4

after simplyfing it equals =n(n+1)(n+2)/6

Answer 2

In the series 1+(1+2)+(1+2+3)....., the nth term is the sum of numbers from 1 to n.
So the nth term is n(n+1)/2.
So we need to find the sum of a series whose nth term is n(n+1)/2.

$\Sigma \frac{n^2}{2}= \frac{n(n+1)(2n+1)}{6*2}=\frac{n(n+1)(2n+1)}{12} \\ \\ \Sigma \frac{n}{2}= \frac{n(n+1)}{2*2} =\frac{n(n+1)}{4} \\ \\ \Sigma \frac{n(n+1)}{2} = \Sigma \frac{n^2+n}{2}= \Sigma \frac{n^2}{2}+ \Sigma \frac{n}{2} \\ \\ \frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4} \\ \\ \frac{n(n+1)}{12}((2n+1)+3)=\frac{n(n+1)}{12} (2n+4) \\ \\ \frac{n(n+1)(n+2)}{6}$