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Question: Find the value of \({{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5..........\left( 2r+1 \right)}}\)...

Find the value of Sn=∑r=1nr1.3.5..........(2r+1){{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5..........\left( 2r+1 \right)}} ?

Explanation

Solution

For answering this question we will simplify and expand this Sn=∑r=1nr1.3.5..........(2r+1){{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5..........\left( 2r+1 \right)}} step by step and cancel the common terms and find the value of Sn{{S}_{n}} . While expanding we will multiply and divide the equation with 2 and add and subtract 1 in the numerator and expand it and observe and cancel all the common terms.

Complete step by step answer:
We need to simplify the given Sn=∑r=1nr1.3.5..........(2r+1){{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5..........\left( 2r+1 \right)}}. We need to find the value of Sn{{S}_{n}}. Let us multiply and divide the given value with 2 after this we will have Sn=12∑r=1n2r1.3.5..........(2r+1){{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\dfrac{2r}{1.3.5..........\left( 2r+1 \right)}} .
Let us add and subtract 1 to the given value after this we will have Sn=12∑r=1n2r+1−11.3.5..........(2r+1){{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\dfrac{2r+1-1}{1.3.5..........\left( 2r+1 \right)}} .
Let us expand this in 2 terms Sn=12∑r=1n(2r+11.3.5..........(2r+1)−11.3.5.......(2r+1)){{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\left( \dfrac{2r+1}{1.3.5..........\left( 2r+1 \right)}-\dfrac{1}{1.3.5.......\left( 2r+1 \right)} \right)} .
Let us cancel the common term (2r+1)\left( 2r+1 \right) in numerator and denominator in the equation we have Sn=12∑r=1n(11.3.5..........(2r−1)−11.3.5.......(2r+1)){{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\left( \dfrac{1}{1.3.5..........\left( 2r-1 \right)}-\dfrac{1}{1.3.5.......\left( 2r+1 \right)} \right)} .
Let us simplify this Sn=12(∑r=1n11.3.5..........(2r−1)−∑r=1n11.3.5..........(2r+1)){{S}_{n}}=\dfrac{1}{2}\left( \sum\limits_{r=1}^{n}{\dfrac{1}{1.3.5..........\left( 2r-1 \right)}}-\sum\limits_{r=1}^{n}{\dfrac{1}{1.3.5..........\left( 2r+1 \right)}} \right) .
Let us expand the equation we have Sn=12[(1+13+13.5+.......+11.3.5.........(2n−1))−(13+13.5+.......+11.3.5.........(2n+1))]{{S}_{n}}=\dfrac{1}{2}\left[ \left( 1+\dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n-1 \right)} \right)-\left( \dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n+1 \right)} \right) \right].
If we observe this all terms are common in both the terms except 2 terms one term in each.
Let us cancel them and have the simplified equation. We will have Sn=12(1−11.3.5.......(2n+1)){{S}_{n}}=\dfrac{1}{2}\left( 1-\dfrac{1}{1.3.5.......\left( 2n+1 \right)} \right) .

Now we will end up with a conclusion having the value of Sn=12(1−11.3.5.......(2n+1)){{S}_{n}}=\dfrac{1}{2}\left( 1-\dfrac{1}{1.3.5.......\left( 2n+1 \right)} \right) for Sn=∑r=1nr1.3.5..........(2r+1){{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5..........\left( 2r+1 \right)}} .

Note: While answering this in the step of expanding the terms we should take care that the expansion comes to be Sn=12[(1+13+13.5+.......+11.3.5.........(2n−1))−(13+13.5+.......+11.3.5.........(2n+1))]{{S}_{n}}=\dfrac{1}{2}\left[ \left( 1+\dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n-1 \right)} \right)-\left( \dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n+1 \right)} \right) \right] not Sn=12[(1+13+13.5+.......+11.3.5.........(2n−1))−(1+13+13.5+.......+11.3.5.........(2n+1))]{{S}_{n}}=\dfrac{1}{2}\left[ \left( 1+\dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n-1 \right)} \right)-\left( 1+\dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n+1 \right)} \right) \right] . If we write this by mistake we will end up having a wrong conclusion as Sn=12(−11.3.5.......(2n+1)){{S}_{n}}=\dfrac{1}{2}\left( -\dfrac{1}{1.3.5.......\left( 2n+1 \right)} \right) . We can observe that this is a completely wrong answer.