1. Mathematical Induction Question

Hi guys, I've been reading the forum for the past 2 semesters of school now and came across a question I couldn't find in the search.

I am to prove 1/1*2*3 +1/2*3*4 + 1/3*4*5 +...+ 1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2)

I am at solving the induction part, but am unable to do finish, I keep getting stuck on what seems to be moderate algebra and I don't know what to do next.

2. Originally Posted by tiddlywinks
Hi guys, I've been reading the forum for the past 2 semesters of school now and came across a question I couldn't find in the search.

I am to prove 1/1*2*3 +1/2*3*4 + 1/3*4*5 +...+ 1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2)

I am at solving the induction part, but am unable to do finish, I keep getting stuck on what seems to be moderate algebra and I don't know what to do next.

I assume you've confirmed the expression is true for n = 1.

assume true for n ... show true for n+1

1/1*2*3 +1/2*3*4 + 1/3*4*5 +...+ 1/n(n+1)(n+2) + 1/(n+1)(n+2)(n+3) =

n(n+3)/[4(n+1)(n+2)] + 1/[(n+1)(n+2)(n+3)] =

n(n+3)^2/[4(n+1)(n+2)(n+3)] + 4/[4(n+1)(n+2)(n+3)] =

[n(n+3)^2 + 4]/[4(n+1)(n+2)(n+3)] =

[n^3 + 6n^2 + 9n + 4]/[4(n+1)(n+2)(n+3)] =

[(n+1)^2 * (n+4)]/[4(n+1)(n+2)(n+3)] =

(n+1)(n+4)/[4(n+2)(n+3)] =

(n+1)[(n+1)+3]/[4[(n+1)+1][(n+1)+2]]

3. Thanks skeeter. I kept screwing up the n^3 part. Guess it was easy algebra and not moderate!