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  4. The sum of n terms of the series(1/1⋅3⋅5) +(1/3⋅5⋅7) +(1/5⋅7⋅9)+ ....is

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Q. The sum of $n$ terms of the series$\frac{1}{1\cdot3\cdot5} +\frac{1}{3\cdot5\cdot7} +\frac{1}{5\cdot7\cdot9}+ ....$is

Sequences and Series

Solution:

If $a_{1}, a_{2}, a_{3} ... a_{n} \in A.P$., then $S_{n}$ of the series
$= \frac{1}{a_{1} a_{2}...a_{r-1} a_{r}} + \frac{1}{a_{2} a_{3}....a_{r+1}}+ ...+\frac{1}{a_{n} a_{n +1}...a_{n +r-1}}$
$= \frac{1}{\left(r-1\right) \left(a_{2}-a_{1}\right)} \left[ \frac{1}{a_{1} a_{2} ...a_{r-1}}-\frac{1}{a_{n +1}...a_{n +r-1}}\right]$
Here $r =3, a_{2} -a_{1} = 3 -1 =2$
$\therefore $ Required Sum $= \frac{1}{2\cdot2}\left[\frac{1}{1\cdot3} -\frac{1}{\left(2n +1\right)\left(2n +3\right)}\right]$
$= \frac{1}{4}\left[\frac{4n^{2} +8n}{3\left(2n +1\right)\left(2n +3\right)}\right] = \frac{n \left(n +2\right)}{3\left(2n +1\right)\left(2n +3\right)}$
Alternative Solution : (Using difference method)
Here $t_{n} = \frac{1}{\left(2n -1\right) \left(2n +1\right)\left(2n +3\right)}$
Let $V_{n} = \frac{1}{\left(2n +1\right)\left(2n +3\right)}$
$\therefore V_{n -1} =\frac{1}{\left(2n -1\right)\left(2n +1\right)}$
Now $V_{n} -V_{n -1} = \frac{1}{\left(2n +1\right)}\left[\frac{1}{2n +3}-\frac{1}{2n -1}\right]$
$= \frac{-4}{\left(2n -1\right)\left(2n +1\right)\left(2n +3\right)}$
$V_{n} -V_{n -1} = - 4t_{n}$
$\therefore t_{n} =\frac{1}{4}\left[V_{n-1}-V_{n}\right]$
Putting $n = 1, 2, 3......,$ then adding
$t_{1} +t_{2} +t_{3} +....t_{n} =S_{n} = \frac{1}{4}\left[V_{0} -V_{n}\right]$
$= \frac{1}{4}\left[\frac{1}{3} -\frac{1}{\left(2n +1\right)\left(2n +3\right)}\right] = \frac{n\left(n +2\right)}{3\left(2n +1\right)\left(2n +3\right)}$
Note: $V_{0}$ is obtained from
$V_{n -1} = \frac{1}{\left(2n -1\right)\left(2n +1\right)}$ by putting $n=1$
$\therefore V_{0} = \frac{1}{3}$