Prove that for every n in the set of natural numbers, N: 11⋅3+13⋅5+15⋅7+…+1(2n−1)⋅(2n+1)=n2n+1

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Answered question

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

Answer & Explanation

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}{(2 n - 1) \cdot (2 n + 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}}{(2 n - 1)} - \frac{\frac{1}{2}}{(2 n + 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}{2 n - 1} - \frac{1}{2 n + 1})

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

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

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

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

Thus the sum of the series is obtained to be

\frac{n}{2 n + 1}

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