( C₀/2 ) - ( C₁/3) + (C₂ /4) - ( C₃/5) +...... =? Pls solve using summation method. Thanks.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{S}=\frac{{}^{\mathrm{n}}{\mathrm{C}}_0}{2}-\frac{{}^{\mathrm{n}}{\mathrm{C}}_1}{3}+\frac{{}^{\mathrm{n}}{\mathrm{C}}_2}{4}-\frac{{}^{\mathrm{n}}{\mathrm{C}}_2}{5}+..... \\ \mathrm{S}=\sum _{\mathrm{r}=0}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}}\frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{\mathrm{r}+2} \\ =\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{ {\left(-1\right)}^{\mathrm{r}} \frac{\left(\mathrm{r}+1\right)}{\left(\mathrm{r}+2\right)}\frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{\left(\mathrm{r}+1\right)}\right\} \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ \frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{\left(\mathrm{r}+1\right)}=\frac{{}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1}}{\left(\mathrm{n}+1\right)} \\ \Rightarrow \mathrm{S}=\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{ {\left(-1\right)}^{\mathrm{r}} \frac{\left(\mathrm{r}+1\right)}{\left(\mathrm{r}+2\right)}\frac{{}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1}}{\left(\mathrm{n}+1\right)}\right\} \\ =\frac{1}{\left(\mathrm{n}+1\right)}\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{ {\left(-1\right)}^{\mathrm{r}} \left(\mathrm{r}+1\right)\frac{{}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1}}{\left(\mathrm{r}+2\right)}\right\} \\ =\frac{1}{\left(\mathrm{n}+1\right)}\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{ {\left(-1\right)}^{\mathrm{r}} \left(\mathrm{r}+1\right)\frac{{}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1}}{\left(\mathrm{r}+1\right)+1}\right\} \\ \mathrm{Again} \mathrm{applying} \mathrm{same} \mathrm{formula} \mathrm{we} \mathrm{get}, \\ \mathrm{S}=\frac{1}{\left(\mathrm{n}+1\right)}\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{ {\left(-1\right)}^{\mathrm{r}} \left(\mathrm{r}+1\right)\frac{{}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}}{\mathrm{n}+2}\right\} \\ =\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{ {\left(-1\right)}^{\mathrm{r}} \left(\mathrm{r}+1\right){.}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}\right\} \\ =\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{ {\left(-1\right)}^{\mathrm{r}} \left(\mathrm{r}+2-1\right){.}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}\right\} \\ =\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}\left[\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{{\left(-1\right)}^{\mathrm{r}} \left(\mathrm{r}+2\right){.}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}\right\}-\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{{\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}\right\}\right] \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{r}{.}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\mathrm{n}{.}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} \\ \Rightarrow \left(\mathrm{r}+2\right){.}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}=\left(\mathrm{n}+2\right){.}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1} \\ \Rightarrow \mathrm{S}=\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}\left[\left(\mathrm{n}+2\right)\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{{\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1}\right\}-\sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{{\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}\right\}\right] \left(\mathrm{i}\right) \\ \sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{{\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1}\right\}{=}^{\mathrm{n}+1}{\mathrm{C}}_1{-}^{\mathrm{n}+1}{\mathrm{C}}_2{+}^{\mathrm{n}+1}{\mathrm{C}}_3-.......... \\ =\left\{{-}^{\mathrm{n}+1}{\mathrm{C}}_0{+}^{\mathrm{n}+1}{\mathrm{C}}_1{-}^{\mathrm{n}+1}{\mathrm{C}}_2{+}^{\mathrm{n}+1}{\mathrm{C}}_3-..........\right\}{+}^{\mathrm{n}+1}{\mathrm{C}}_0 \\ =\left({}^{\mathrm{n}+1}{\mathrm{C}}_1{+}^{\mathrm{n}+1}{\mathrm{C}}_3+...\right)-\left({}^{\mathrm{n}+1}{\mathrm{C}}_0{+}^{\mathrm{n}+1}{\mathrm{C}}_2+...\right){+}^{\mathrm{n}+1}{\mathrm{C}}_0 \\ =0{+}^{\mathrm{n}+1}{\mathrm{C}}_0 \left[\mathrm{Sum} \mathrm{of} \mathrm{odd} \mathrm{Binomial} \mathrm{Coefficient}=\mathrm{Sum} \mathrm{of} \mathrm{Even} \mathrm{Binomial} \mathrm{Coefficient}\right] \\ {=}^{\mathrm{n}+1}{\mathrm{C}}_0=1 \\ \sum _{\mathrm{r}=0}^{\mathrm{n}}\left\{{\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+2}\right\}{=}^{\mathrm{n}+2}{\mathrm{C}}_2{-}^{\mathrm{n}+2}{\mathrm{C}}_3{+}^{\mathrm{n}+2}{\mathrm{C}}_4{-}^{\mathrm{n}+2}{\mathrm{C}}_5+... \\ =\left\{{}^{\mathrm{n}+2}{\mathrm{C}}_0{-}^{\mathrm{n}+2}{\mathrm{C}}_1{+}^{\mathrm{n}+2}{\mathrm{C}}_2{-}^{\mathrm{n}+2}{\mathrm{C}}_3{+}^{\mathrm{n}+2}{\mathrm{C}}_4{-}^{\mathrm{n}+2}{\mathrm{C}}_5+...\right\}{+}^{\mathrm{n}+2}{\mathrm{C}}_1{-}^{\mathrm{n}+2}{\mathrm{C}}_0 \\ =0{+}^{\mathrm{n}+2}{\mathrm{C}}_1{-}^{\mathrm{n}+2}{\mathrm{C}}_0 \\ =\mathrm{n}+2-1 \\ =\mathrm{n}+1 \\ \mathrm{Hence} \left(\mathrm{i}\right) \mathrm{becomes}: \\ \mathrm{S}=\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}\left[\left(\mathrm{n}+2\right)-\left(\mathrm{n}+1\right)\right] \\ =\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}\left[\mathrm{n}+2-\mathrm{n}-1\right] \\ =\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)} \\ \Rightarrow \boxed{\frac{{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{-}\frac{{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{1}}}{\mathbf{3}}\mathbf{+}\frac{{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{2}}}{\mathbf{4}}\mathbf{-}\frac{{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{2}}}{\mathbf{5}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{=}\frac{\mathbf{1}}{\left(\mathbf{n}\mathbf{+}\mathbf{1}\right)\left(\mathbf{n}\mathbf{+}\mathbf{2}\right)}} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards