Finding the radius of a cylinder

**igcsestudent** · May 14th 2011, 09:16 AM

Hello,
can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
good luck
thanks

**HallsofIvy** · May 14th 2011, 11:48 AM

$S_n$ 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:
$S_{n}- S_k{n-1}= \frac{1}{6}(n)(n+1)(n+2- (n-1))= \frac{1}{6}n(n+1)(3)$

**bjhopper** · May 14th 2011, 12:14 PM

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

**igcsestudent** · May 15th 2011, 03:05 AM

cool!
thanks really appreciate your help!

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