Question 1104186
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Each of the fractions is of the form {{{1/(n(n+2))}}}.<br>
Whenever you see a problem like this, where you are adding several terms of the same form, it is almost certain that you can write each term in such a way that, when all the terms are added, most of the terms cancel, leaving you with a relatively simple answer.<br>
So let's find a way to "decompose" the general term into the sum of two terms.  Since the denominator is n(n+2), we want to write the term as the sum of two fractions, one with denominator n and the other with denominator (n+2).<br>
{{{1/(n(n+2)) = A/n + B/(n+2)}}}
{{{1/n(n+2) = (A(n+2) + B(n))/(n(n+2))}}}
{{{1/n(n+2) = ((A+B)n + 2A)/(n(n+2))}}}
{{{1 = (A+B)n + 2A}}}<br>
This equation must be true for all values of n; that means
{{{A+B = 0}}}  and  {{{2A = 1}}}<br>
Solving that system of equations gives us
A = 1/2; B = -1/2<br>
So the general term of the sequence can be written as
{{{1/(n(n+2)) = (1/2)/n - (1/2)/(n+2) = (1/2)(1/n-1/(n+2))}}}<br>
Now rewrite each term as the sum of two terms, using this pattern:<br>
{{{1/2(4) = (1/2)(1/2-1/4)}}}
{{{1/3(5) = (1/2)(1/3-1/5)}}}
{{{1/4(6) = (1/2)(1/4-1/6)}}}
{{{1/5(7) = (1/2)(1/5-1/7)}}}
...
{{{1/11(13) = (1/2)(1/11-1/13)}}}
{{{1/12(14) = (1/2)(1/12-1/14)}}}
{{{1/13(15) = (1/2)(1/13-1/15)}}}<br>
Now looking at the terms in parentheses, you see that there is a 1/4 and a -1/4, a 1/5 and a -1/5, ..., and a 1/13 and a -1/13.<br>
So all those terms cancel; all that is left is the first fractions from the original first two terms and the second fractions from the original last two terms:
{{{(1/2)(1/2+1/3-1/14-1/15)}}}<br>
So, using a common denominator, the sum we are looking for is<br>
{{{(1/2)(105/210+70/210-15/210-14/210) = (1/2)(146/210) = 73/210}}}