Hello,
can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
good luck
thanks
$\displaystyle S_n$ is defined by $\displaystyle S_n= \frac{1}{6}n(n+1)(n+2)$ so $\displaystyle S_{n-1}= \frac{1}{6}(n-1)((n-1)+ 1)((n-1)+ 2)= \frac{1}{6}(n-1)(n)(n+1)$
You could multiply both of those out and subtract but it is simpler to notice that each has factors of 1/6, n, and n+1 andfactor those out:
$\displaystyle S_{n}- S_k{n-1}= \frac{1}{6}(n)(n+1)(n+2- (n-1))= \frac{1}{6}n(n+1)(3)$