Find the Sum

$\displaystyle \sum_{n=1}^{91} {\frac{1}{\sqrt{n+8}+\sqrt{n+9}}} = \sum_{n=9}^{99} {\frac{1}{\sqrt{n}+\sqrt{n+1}}}$

$\displaystyle = \sum_{n=9}^{99} {\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}} = \sum_{n=9}^{99} {\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}}$

$\displaystyle = \sum_{n=9}^{99} {(\sqrt{n+1}-\sqrt{n})} = \sum_{n=10}^{100} {\sqrt{n}} - \sum_{n=9}^{99} {\sqrt{n}} = \sqrt{100} - \sqrt{9} = 10 - 3 = \boxed{7}$

By rationalising we can get this series as

$\dfrac{\sqrt{10}-\sqrt{9}}{1}+\dfrac{\sqrt{11}-\sqrt{10}}{1}........\dfrac{\sqrt{10+(n-1)}-\sqrt{9+(n-1)}}{1}$

We have to sum up till $n=91$

$\sqrt{10}-\sqrt{9}+\sqrt{11}-\sqrt{10}..........+\sqrt{100}-\sqrt{99}$

every term will cancel out except $\sqrt9$ and $\sqrt{100}$

which will give us $10-3=7$ .

Python:

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from math import sqrt
n = 9
terms = 0
total = 0
while terms != 91:
    total += 1/(sqrt(n)+sqrt(n+1))
    n += 1
    terms += 1
print "Answer:", total