# Method of differences exam question

1. Nov 30, 2008

### thomas49th

1. The problem statement, all variables and given/known data

given the identity $$\frac{2}{(r+1)(r+3)} \equiv \frac{1}{r+1} - \frac{1}{r+3}$$

prove that

$$\sum^{n}_{r = 1} \frac{2}{(r+1)(r+3)} = \frac{n(an + b)}{6(n+2)(n+3)}$$

where a and b are constants to be found

2. Relevant equations

3. The attempt at a solution

I subed in number r = 1,2,3,n

r =1: 1/2 - 1/4
r=2:1/3 - 1/5
r= 3: 1/4-1/6
r = n: 1/(n+1) - 1/(n+3)

the ones in red being the ones that dont cancel... so i add them but i get 5n-9/6(n+3)

is there somthing else i need to do?

Last edited: Nov 30, 2008
2. Nov 30, 2008

### lurflurf

you are making this hard
given the partial fractions hint
write out the inverses of the sums posive one row ngative the next
2,3,4,5,6,7,8,9,10...
4,5,6,7,8,9,10,11,12,13...
so past the first few terms we net 2 positive and 2 negative terms
sum=1/2+1/3-1/(n+2)-1/(n+3)

in short you forgot -1/(n+2) does not cancel

Last edited: Nov 30, 2008