tinfoQ

2021-10-09

Prove that for every n in the set of natural numbers, N:

$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots +\frac{1}{(2n-1)\cdot (2n+1)}=\frac{n}{2n+1}$

Latisha Oneil

Skilled2021-10-10Added 100 answers

Step 1

The partial fraction is the process of splitting the terms in the denominator into different fractions. This can be used to simplify the fraction for further evaluation.

Step 2

The series is evaluated by splitting the terms into partial fractions as follows:-

$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots +\frac{1}{(2n-1)\cdot (2n+1)}=(\frac{\frac{1}{2}}{1}-\frac{\frac{1}{2}}{3})+(\frac{\frac{1}{2}}{3}-\frac{\frac{1}{2}}{5})+(\frac{\frac{1}{2}}{5}-\frac{\frac{1}{2}}{7})+\dots (\frac{\frac{1}{2}}{(2n-1)}-\frac{\frac{1}{2}}{(2n+1)})$

$=\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\dots +\frac{1}{2n-1}-\frac{1}{2n+1})$

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

$=\frac{1}{2}\left(\frac{2n+1-1}{2n+1}\right)$

$=\frac{1}{2}\left(\frac{2n}{2n+1}\right)$

$=\frac{n}{2n+1}$

Thus the sum of the series is obtained to be$\frac{n}{2n+1}$

The partial fraction is the process of splitting the terms in the denominator into different fractions. This can be used to simplify the fraction for further evaluation.

Step 2

The series is evaluated by splitting the terms into partial fractions as follows:-

Thus the sum of the series is obtained to be

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