Q 13 pls help.
Q13. If $3.{}^{n}C_0-8.{}^{n}C_1+13.{}^{n}C_2-18.{}^{n}C_3+.....$ upto $\left(n+1\right)$ terms is $\left(n\ge 2\right)$:
(A) zero
(B) 1
(C) 2
(D) none of these

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

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \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} \left(\mathrm{n}+1\right) \mathrm{terms} \\ 3,8,13, \mathrm{is} \mathrm{an} \mathrm{A}.\mathrm{P}. \mathrm{with} \mathrm{common} \mathrm{difference} \mathrm{of} \\ \mathrm{Hence} \\ {\mathrm{T}}_{\mathrm{r}}=\left(5\mathrm{r}+3\right){ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} {\left(-1\right)}^{\mathrm{r}} \mathrm{where} \mathrm{r}\ge 0 \\ \mathrm{Hence} \\ \mathrm{S}=\sum _{\mathrm{r}=0}^{\mathrm{n}} \left(5\mathrm{r}+3\right){ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} {\left(-1\right)}^{\mathrm{r}} \\ =5\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} {\left(-1\right)}^{\mathrm{r}}+3\sum _{\mathrm{r}=0}^{\mathrm{n}} { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} {\left(-1\right)}^{\mathrm{r}} \\ \mathrm{Series} \sum _{\mathrm{r}=0}^{\mathrm{n}} { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} {\left(-1\right)}^{\mathrm{r}} \mathrm{goes} \mathrm{from}{ }^{\mathrm{n}}{\mathrm{C}}_0 \mathrm{to}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}} \left(\mathrm{Complete} \mathrm{series}\right)\mathrm{with} \mathrm{alternate} \mathrm{positive} \mathrm{and} \mathrm{negative} \\ \mathrm{sign} , \mathrm{hence} \sum _{\mathrm{r}=0}^{\mathrm{n}} { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} {\left(-1\right)}^{\mathrm{r}}=0 \\ \mathrm{S}=5\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}} {\left(-1\right)}^{\mathrm{r}}+0 \\ \mathrm{Use} \mathrm{r}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\mathrm{n}{ }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} \\ \mathrm{S}=5\mathrm{n} \sum _{\mathrm{r}=1}^{\mathrm{n}} { }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} {\left(-1\right)}^{\mathrm{r}} \\ =-5\mathrm{n} \sum _{\mathrm{r}=1}^{\mathrm{n}} { }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} {\left(-1\right)}^{\mathrm{r}-1} \\ \mathrm{Now} \mathrm{series} \sum _{\mathrm{r}=1}^{\mathrm{n}} { }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} {\left(-1\right)}^{\mathrm{r}-1} \mathrm{goes} \mathrm{from}{ }^{\mathrm{n}-1}{\mathrm{C}}_0 \mathrm{to}{ }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}-1} \mathrm{with} \\ \mathrm{alternate} \mathrm{positive} \mathrm{and} \mathrm{negative} \mathrm{sign}, \mathrm{hence} \sum _{\mathrm{r}=1}^{\mathrm{n}} { }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1} {\left(-1\right)}^{\mathrm{r}-1}=0 \\ \Rightarrow \mathrm{S}=-5\mathrm{n}\times 0=0 \\ 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} \left(\mathrm{n}+1\right) \mathrm{terms}=0 \left[\mathrm{Answer}\right] \end{gathered}$$

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