text: 1. ## Algebra
  
  Hello,
  can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
  good luck
  thanks
  
  2. $S_n$ is defined by $S_n= \frac{1}{6}n(n+1)(n+2)$ so $S_{n-1}= \frac{1}{6}(n-1)((n-1)+ 1)((n-1)+ 2)= \frac{1}{6}(n-1)(n)(n+1)$
  
  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:
  $S_{n}- S_k{n-1}= \frac{1}{6}(n)(n+1)(n+2- (n-1))= \frac{1}{6}n(n+1)(3)$
  
  3. ## sequence oroblem
  
  Originally Posted by igcsestudent
  Hello,
  can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
  good luck
  thanks
  sequence 1,4,10,20,35 Find nextterm
  Differences 3,6,10, 15, next difference is 21 6 comes after 3,4,5
  35+21 is 56
  
  bjh
  
  4. cool!