SOLUTION: Use mathematical induction to prove that: n ∑ r^2/{(2r-1)(2r+1)}= n(n+1)/(2(2n+1)) r=1 Thanks!!

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Question 941989: Use mathematical induction to prove that:
n
∑ r^2/{(2r-1)(2r+1)}= n(n+1)/(2(2n+1))
r=1
Thanks!!
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
We first want to prove that
if equation (1) were true for some value of n,

(1) %22%22=%22%22n%28n%2B1%29%2F%282%282n%2B1%29%29 then it would also be true when we replace n by n+1. That is,
we want to show that if (1) holds for some value of n, then we
can erase the ??? below and replace it with an equal sign:
%22%3F%3F%3F%22
which simplifies to:
(2) %22%3F%3F%3F%22%28n%2B1%29%28n%2B2%29%2F%282%282n%2B3%29%29
-----------------------
So we add the (n+1)st term to both sides of (1)
%22%22%2B%22%22 %28n%2B1%29%5E2%2F%28%282%28n%2B1%29%5E%22%22-1%29%282%28n%2B1%29%5E%22%22%2B1%29%29 %22%22=%22%22 n%28n%2B1%29%5E%22%22%2F%282%282n%2B1%29%5E%22%22%29 %22%22%2B%22%22 %28n%2B1%29%5E2%2F%28%282%28n%2B1%29%5E%22%22-1%29%282%28n%2B1%29%5E%22%22%2B1%29%29
and simplify. The left side is just the sum to n+1 terms instead of n
terms:
%22%22=%22%22 n%28n%2B1%29%5E%22%22%2F%282%282n%2B1%29%5E%22%22%29 %22%22%2B%22%22 %28n%2B1%29%5E2%2F%282n%2B1%29%282n%2B3%29%29
Get an LCD by multiplying the first fraction on the right by %28%282n%2B3%29%29%2F%28%282n%2B3%29%29 and the second by 2%2F2
%22%22=%22%22 n%28n%2B1%29%5E%22%22%282n%2B3%29%2F%282%282n%2B1%29%5E%22%22%282n%2B3%29%29 %22%22%2B%22%22 2%28n%2B1%29%5E2%2F2%282n%2B1%29%282n%2B3%29%29
%22%22=%22%22 n%28n%2B1%29%5E%22%22%282n%2B3%29%2F%282%282n%2B1%29%5E%22%22%282n%2B3%29%29 %22%22%2B%22%22 2%28n%2B1%29%5E2%2F2%282n%2B1%29%282n%2B3%29%29
Combine numerators over the LCD
%22%22=%22%22
Factor out (n+1) on top
%22%22=%22%22
%22%22=%22%22 %28%28n%2B1%29%282n%5E2%2B3n%2B2n%2B2%29%29%2F%282%282n%2B1%29%5E%22%22%282n%2B3%29%29
%22%22=%22%22 %28%28n%2B1%29%282n%5E2%2B5n%2B2%29%29%2F%282%282n%2B1%29%5E%22%22%282n%2B3%29%29
%22%22=%22%22 %28%28n%2B1%29%282n%2B1%29%28n%2B2%29%5E%22%22%29%2F%282%282n%2B1%29%5E%22%22%282n%2B3%29%29
Now cancel the (2n+1)'s
%22%22=%22%22 %28%28n%2B1%29%28n%2B2%29%5E%22%22%29%2F%282%282n%2B3%29%29
That is the same as (2)
All that's left is to show that (1) is true for n=1

1%5E2%2F%28%282%281%29%5E%22%22-1%29%282%281%29%5E%22%22%2B1%29++++++++%29%22%22=%22%221%281%2B1%29%2F%282%282%281%29%5E%22%22%2B1%29%29
1%2F%28%282-1%29%282%2B1%29++++++++%29%22%22=%22%221%282%29%2F%282%282%2B1%29%29
1%2F%28%281%29%283%29++++++++%29%22%22=%22%222%2F%282%283%29%29
1%2F3 %22%22=%22%22 1%2F3
So since it works for n=1, it works for n=2. Then
since it works for n=2, it works for n=3, then for n=4
and so on forever for all values on n.
Edwin