Question

$1+3+6+10+...+\frac{(n-1)n}{2}+\frac{n(n+1)}{2}=$

• A. $\frac{n(n+1)(n+2)}{3}$
• B. $\frac{(n+1)(n+2)}{6}$
• C. $\frac{n(n+1)(n+2)}{6}$
• D. $\frac{(n+2)(n+1{)}^2}{3}$

Answer 1

Answer: (C)

$\frac{n(n+1)(n+2)}{6}$

We can see that $T_n=\frac{n(n+1)}{2}$
hence $\Rightarrow S_n={\sum }_{n=1}^n{T}_n\Rightarrow S_n={\sum }_{n=1}^n\frac{n(n+1)}{2}=\frac{1}{2}({\sum }_{n=1}^n{n}^2+{\sum }_{n=1}^{n}n)=\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}]=\frac{1}{2}\left[\frac{n(n+1)(2n+1+3)}{6}\right]=\frac{2}{2}\left[\frac{n(n+1)(n+2)}{6}\right]=\frac{n(n+1)(n+2)}{6}$