Question 1104112
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Let's look at the sum of the first few terms of the series; then you can use that to predict the sum of the whole series.<br>
{{{ 5/(sqrt (2) + sqrt (1)) =(5/(sqrt (2) + sqrt (1)))*((sqrt(2)-sqrt(1))/(sqrt(2)-sqrt(1))) = 5*((sqrt(2)-sqrt(1))/(2-1)) = 5*(sqrt(2)-sqrt(1))}}}<br>
{{{ 5/(sqrt (3) + sqrt (2)) =(5/(sqrt (3) + sqrt (2)))*((sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2))) = 5*((sqrt(3)-sqrt(2))/(3-2)) = 5*(sqrt(3)-sqrt(2))}}}<br>
{{{ 5/(sqrt (4) + sqrt (3)) =(5/(sqrt (4) + sqrt (3)))*((sqrt(4)-sqrt(3))/(sqrt(4)-sqrt(3))) = 5*((sqrt(4)-sqrt(3))/(4-3)) = 5*(sqrt(4)-sqrt(3))}}}<br>
So the sum of the first three terms of the series is<br>
{{{5*(sqrt(2)-sqrt(1))+5*(sqrt(3)-sqrt(2))+5*(sqrt(4)-sqrt(3)) = 5*((sqrt(2)-sqrt(1))+(sqrt(3)-sqrt(2))+(sqrt(4)-sqrt(3))) = 5*(sqrt(4)-sqrt(1))}}}<br>
You can see that the terms with sqrt(2) and sqrt(3) cancelled, leaving only sqrt(4) and sqrt(1).<br>
The last term in the series is going to be equivalent to {{{5*(sqrt(36)-sqrt(35))}}}<br>
Can you see what you are going to be left with when you add all the terms of the series and cancel where possible?