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Mathematics Question on Sequence and series

Find the sum to nn terms of the series 11⋅3+13⋅5+15⋅7+...\frac{1}{1\cdot3} +\frac{1}{3\cdot5}+\frac{1}{5\cdot7} +...

A

n(2n−1)\frac{n}{\left(2n-1\right)}

B

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

C

n+1n+1

D

n−1n-1

Answer

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

Explanation

Solution

Let TrT_{r} be the rthr^{th} term of the given series. Then, Tr=1(2r−1)(2r+1),r=1,2,3,...,nT_{r} = \frac{1}{ \left(2r-1\right)\left(2r+1\right)}, r= 1, 2, 3, ..., n ⇒Tr=12(12r−1−12r+1)\Rightarrow T_{r} = \frac{1}{2} \left(\frac{1}{2r-1} - \frac{1}{ 2r+1}\right) ∴\therefore Required sum =∑r=1nTr=12[(11−13)+(13−15)+(15−17)+...+(12n−1−12n+1)] = \sum\limits_{r=1}^{n}T_{r} = \frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{2n-1} -\frac{1}{2n+1}\right)\right] =12[1−12n+1]=n2n+1 = \frac{1}{2}\left[1-\frac{1}{2n+1}\right] = \frac{n}{2n+1}