Question

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

Answer 1

Hint: For answering this question we will simplify and expand this ${{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5..........\left( 2r+1 \right)}}$ step by step and cancel the common terms and find the value of ${{S}_{n}}$ . 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 ${{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{r}{1.3.5..........\left( 2r+1 \right)}}$. We need to find the value of ${{S}_{n}}$. Let us multiply and divide the given value with 2 after this we will have ${{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\dfrac{2r}{1.3.5..........\left( 2r+1 \right)}}$ .
Let us add and subtract 1 to the given value after this we will have ${{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\dfrac{2r+1-1}{1.3.5..........\left( 2r+1 \right)}}$ .
Let us expand this in 2 terms ${{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\left( \dfrac{2r+1}{1.3.5..........\left( 2r+1 \right)}-\dfrac{1}{1.3.5.......\left( 2r+1 \right)} \right)}$ .
Let us cancel the common term $\left( 2r+1 \right)$ in numerator and denominator in the equation we have ${{S}_{n}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\left( \dfrac{1}{1.3.5..........\left( 2r-1 \right)}-\dfrac{1}{1.3.5.......\left( 2r+1 \right)} \right)}$ .
Let us simplify this ${{S}_{n}}=\dfrac{1}{2}\left( \sum\limits_{r=1}^{n}{\dfrac{1}{1.3.5..........\left( 2r-1 \right)}}-\sum\limits_{r=1}^{n}{\dfrac{1}{1.3.5..........\left( 2r+1 \right)}} \right)$ .
Let us expand the equation we have ${{S}_{n}}=\dfrac{1}{2}\left[ \left( 1+\dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n-1 \right)} \right)-\left( \dfrac{1}{3}+\dfrac{1}{3.5}+.......+\dfrac{1}{1.3.5.........\left( 2n+1 \right)} \right) \right]$.
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 ${{S}_{n}}=\dfrac{1}{2}\left( 1-\dfrac{1}{1.3.5.......\left( 2n+1 \right)} \right)$ .

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

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