Method of differences exam question

[thomas49th]

Homework Statement

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

Homework Equations

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 don't cancel... so i add them but i get 5n-9/6(n+3)

is there somthing else i need to do?

 

[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