Partial sum of a series

For 1/(k(k+2)) you have 3/4 - (2n+3)/(2(n+1)(n+2)). How do you get there from 1/(k(k+2))? Having trouble getting from a to b.

If you use a partial fraction decomposition, it becomes clear what is going on with this sum. Note that

1/(k(k+2)) = ½/k - ½/(k + 2).

So, forgetting about the 1/2 for a moment, look at the sum

n
Σ (1/k - 1/(k+2))
k=1

This is telescoping. If we write out a bunch of terms, we can see that stuff cancels:

(1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + ... + (1/(n-2) - 1/n) + (1/(n-1) - 1/(n+1)) + (1/n - 1/(n+2)).

Okay, so the 1 and 1/2 stay, and the terms -1/(n+1) and -1/(n+2) stay. Everything else cancels with it's like term.

Putting back the factor of 1/2, the partial sum is

½(1 + 1/2 - 1/(n+1) - 1/(n+2)) = 3/4 - (1/2) (2n+3)/((n+1)(n+2)).

The PFs are (1/2)/(k) - (1/2)/(k+2) and its good to extend,
=[(1/2)/(k) - (1/2)/(k+1)] + [(1/2)/(k+1) - (1/2)/(k+2)]
Sum each square bracket fro k=1 to n
[(1/2) - 1/(2(n+1) + [1/4 - 1/(2(n+2)] and you get the required answer.
Note that the sum of [f(k) - f(k+1)] from k=1 to n is f(1) - f(n+1).