# Polynomials

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## Sum of fractions

${\frac{3}{4} + \frac{5}{36} + \frac{7}{144} + \frac{9}{400} +.... + \frac{19}{8100}}$ = ?
1. ${\frac{9}{10}}$
2. ${\frac{11}{18}}$
3. ${\frac{99}{100}}$
4. ${\frac{80}{81}}$

Choice C. ${\frac{99}{100}}$

## Detailed Solution

${\frac{3}{2^2} + \frac{5}{6^2} + \frac{7}{12^2} + \frac{9}{20^2} ... \frac{19}{90^2} }$

So, ${t_n = \frac{(2n+1)}{((n)^2(n+1)^2)} }$

Now, we are back to the partial fractions approach,

${t_n = \frac{(2n+1)}{((n)^2(n+1)^2)} = \frac{1}{(n)^2} - \frac{1}{(n+1)^2} }$

So, the above expression is nothing but

${t_n = (\frac{1}{1^2} - \frac{1}{2^2}) + (\frac{1}{2^2} - \frac{1}{3^2}) + (\frac{1}{3^2} - \frac{1}{4^2}) + (\frac{1}{4^2} - \frac{1}{5^2}) ... + (\frac{1}{9^2} - \frac{1}{10^2}) = (\frac{1}{1^2} - \frac{1}{10^2}) }$

= ${\frac{99}{100} }$

Correct Answer: ${\frac{99}{100}}$.

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