Please help me with 13th sum
correct option is A

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

$$\begin{gathered} \mathrm{S}=3{.}^{\mathrm{n}}{\mathrm{C}}_0-8{.}^{\mathrm{n}}{\mathrm{C}}_1+13{.}^{\mathrm{n}}{\mathrm{C}}_2-18{.}^{\mathrm{n}}{\mathrm{C}}_3+..\mathrm{upto} \mathrm{n}+1 \mathrm{terms} \\ 3,8,13 \mathrm{is} \mathrm{an} \mathrm{A}.\mathrm{P}. \mathrm{of} \mathrm{first} \mathrm{term} \mathrm{and} \mathrm{common} \mathrm{difference} 5 \\ {\mathrm{r}}_{\mathrm{th}} \mathrm{term} \mathrm{is} \mathrm{given} \mathrm{by}: \\ {\mathrm{T}}_{\mathrm{r}}=3+\left(\mathrm{r}-1\right)5=5\mathrm{r}-2 \\ \mathrm{But} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{start} \mathrm{from} \mathrm{r}=0 \\ 5\mathrm{r}-2+5=5\mathrm{r}+3 \\ \mathrm{From} \mathrm{r}=0 \mathrm{to} \mathrm{r}=\mathrm{n}, \mathrm{there} \mathrm{will} \mathrm{be} \mathrm{n}+1 \mathrm{terms} \\ \mathrm{S}=\sum _{\mathrm{r}=0}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} \left(5\mathrm{r}+3\right){.}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} \\ =5\sum _{\mathrm{r}=0}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} \mathrm{r}{.}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}+3\sum _{\mathrm{r}=0}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} \\ \mathrm{S}=5\sum _{\mathrm{r}=1}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} \mathrm{n}{.}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1}+3\sum _{\mathrm{r}=0}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} ....\left(\mathrm{i}\right)\left[\mathrm{As} \mathrm{r}{.}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\mathrm{n}{.}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1}\right] \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ {\left(1+\mathrm{x}\right)}^{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0{+}^{\mathrm{n}}{\mathrm{C}}_1\mathrm{x}{+}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{x}}^2+....+{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}} {\mathrm{x}}^{\mathrm{n}} \\ \mathrm{Put} \mathrm{x}=-1 \\ \Rightarrow {\left(1-1\right)}^{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0{-}^{\mathrm{n}}{\mathrm{C}}_1{+}^{\mathrm{n}}{\mathrm{C}}_2-....+{\left(-1\right)}^{\mathrm{n}}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}} \\ \Rightarrow 0=\sum _{\mathrm{r}=0}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} {.}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} \\ \mathrm{i}.\mathrm{e}. \\ \mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }{\left(\mathbf{-}\mathbf{1}\right)}^{\mathbf{r}}\mathbf{ }{\mathbf{.}}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{.}\mathbf{.}\left(\mathbf{ii}\right) \\ {\left(1+\mathrm{x}\right)}^{\mathrm{n}-1}{=}^{\mathrm{n}-1}{\mathrm{C}}_0{+}^{\mathrm{n}-1}{\mathrm{C}}_1\mathrm{x}{+}^{\mathrm{n}-1}{\mathrm{C}}_2{\mathrm{x}}^2+...{+}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}} \\ \mathrm{Put} \mathrm{x}=-1 \\ \Rightarrow {\left(1-1\right)}^{\mathrm{n}-1}{=}^{\mathrm{n}-1}{\mathrm{C}}_0{-}^{\mathrm{n}-1}{\mathrm{C}}_1{+}^{\mathrm{n}-1}{\mathrm{C}}_2+...+{\left(-1\right)}^{\mathrm{n}}{ }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}-1} \\ \Rightarrow {\left(1-1\right)}^{\mathrm{n}-1}=\sum _{\mathrm{r}=1}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} \mathrm{n}{.}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} \\ \Rightarrow 0=\sum _{\mathrm{r}=1}^{\mathrm{n}} {\left(-1\right)}^{\mathrm{r}} \mathrm{n}{.}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} \\ \Rightarrow \mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\mathbf{ }{\left(\mathbf{-}\mathbf{1}\right)}^{\mathbf{r}}\mathbf{ }\mathbf{n}{\mathbf{.}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{C}}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{0}...\left(\mathrm{iii}\right) \\ \mathrm{Using} \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right) \mathrm{in} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get}: \\ \mathrm{S}=5\times 0+3\times 0=\mathbf{0}\mathbf{\to }\mathbf{Option}\left(\mathbf{A}\right) \end{gathered}$$

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