John,
You might first look at the typical term 1/n(n+2) and think about rewriting it in terms of 1/n and 1/(n+2); for example
^{1}/_{1x3} = ( ^{1}/_{2})( ^{1}/_{1}  ^{1}/_{3}),
^{1}/_{2x4} = ( ^{1}/_{2})( ^{1}/_{2}  ^{1}/_{4}), etc.
Penny
John wrote back.
Hi,
I still don't quiet understand this question you answered for me. The answer you gave me for the formula is the same formula that as in the question and according to my son you can't have the same formula, it can be similar but not the same. So if you could help me out and figure this question out for me I would greatly appreciate it alot.
John
John,
Using the technique I hinted at above you can develop an expression for the sum 1/(1x3)+1/(2x4)+1/(3x5)...+1/(n(n+2)). For example if n = 5
1/(1x3)+1/(2x4)+1/(3x5)...+1/(5(5+2))
= 1/2(1/1  1/3) + 1/2(1/2  1/4) + 1/2(1/3  1/5)
+ 1/2(1/4  1/6) + 1/2(1/5  1/7)
If you examine this expression you will see there is some cancellation and the expression reduces to
1/(1x3)+1/(2x4)+1/(3x5)...+1/(5(5+2))
= 1/2(1/1 + 1/2  1/6  1/7)
Apply this same logic to 1/(1x3)+1/(2x4)+1/(3x5)...+1/(n(n+2)) in order to find an expression for the sum.
Penny
