Question

The value of sum of the series: $3·{}^{n}C_0-8·{}^{n}C_1+13·{}^{n}C_2-18·{}^{n}C_3+......$up to $\left(n+1\right)$ terms is

• A. $0$
• B. $3n$
• C. $5n$
• D. None of these

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Step 1. Find the $r^{th}$ term of the series

The given series: $3·{}^{n}C_0-8·{}^{n}C_1+13·{}^{n}C_2-18·{}^{n}C_3+......$up to $\left(n+1\right)$ terms.

The general term of the series can be given by, $t_r={\left(-1\right)}^r\left(3+5r\right){}^{n}C_r;r=0,1,2,....,n$.

$$\begin{gathered} \Rightarrow t_r={\left(-1\right)}^r\left(3\right)\left({}^{n}C_r\right)+{\left(-1\right)}^r\left(5r\right){}^{n}C_r \\ \Rightarrow t_r={\left(-1\right)}^r\left(3\right)\left({}^{n}C_r\right)+{\left(-1\right)}^r\left(5r\right)\frac{n!}{r!\times \left(n-r\right)!} \\ \Rightarrow t_r={\left(-1\right)}^r\left(3\right)\left({}^{n}C_r\right)+{\left(-1\right)}^r\left(5r\right)\frac{n\times \left(n-1\right)!}{r\times \left(r-1\right)!\times \left[\left(n-1\right)-\left(r-1\right)\right]!} \\ \Rightarrow t_r={\left(-1\right)}^r\left(3\right)\left({}^{n}C_r\right)+{\left(-1\right)}^r\left(5r\right)\left(\frac{n}{r}\right)\frac{\left(n-1\right)!}{\left(r-1\right)!\times \left[\left(n-1\right)-\left(r-1\right)\right]!} \\ \Rightarrow t_r={\left(-1\right)}^r\left(3\right)\left({}^{n}C_r\right)+{\left(-1\right)}^r\left(5n\right)\left({}^{n-1}C_{r-1}\right) \end{gathered}$$

Step 2. Find the expression of the sum

Thus, the sum of the series can be given by, $S=3·{}^{n}C_0-8·{}^{n}C_1+13·{}^{n}C_2-18·{}^{n}C_3+......$up to $\left(n+1\right)$ terms.

$$\begin{gathered} \Rightarrow S=\sum _{r=0}^n{t}_r \\ \Rightarrow S=\sum _{r=0}^n\left[{\left(-1\right)}^r\left(3\right)\left({}^{n}C_r\right)+{\left(-1\right)}^r\left(5n\right)\left({}^{n-1}C_{r-1}\right)\right] \\ \Rightarrow S=\sum _{r=0}^n\left[{\left(-1\right)}^r\left(3\right)\left({}^{n}C_r\right)\right]+\sum _{r=0}^n\left[{\left(-1\right)}^r\left(5n\right)\left({}^{n-1}C_{r-1}\right)\right] \\ \Rightarrow S=\left(3\right)\sum _{r=0}^n\left[{\left(-1\right)}^r\left({}^{n}C_r\right)\right]+\left(5n\right)\sum _{r=0}^n\left[{\left(-1\right)}^r\left({}^{n-1}C_{r-1}\right)\right] \end{gathered}$$

Step 3. Find the value of the sum

From the binomial expansion theorem, ${\left(1+x\right)}^n=\sum _{r=0}^n{}^{n}C_r{\left(1\right)}^{n-r}{\left(x\right)}^r$

Put, $x=-1$.

$$\begin{gathered} \Rightarrow {\left(1-1\right)}^n=\sum _{r=0}^n{}^{n}C_r{\left(-1\right)}^r \\ \Rightarrow \sum _{r=0}^n{\left(-1\right)}^r\left({}^{n}C_r\right)=0 \end{gathered}$$

From the binomial expansion theorem, ${\left(1+x\right)}^{n-1}=\sum _{r=1}^n{}^{n-1}C_{r-1}{\left(1\right)}^{n-r}{\left(x\right)}^r$

Put, $x=-1$.

$$\begin{gathered} \Rightarrow {\left(1-1\right)}^n=\sum _{r=1}^n\left({}^{n-1}C_{r-1}\right){\left(-1\right)}^{r-1} \\ \Rightarrow \sum _{r=0}^n{\left(-1\right)}^r\left({}^{n-1}C_{r-1}\right)=0 \end{gathered}$$

Consider the expression: $S=\left(3\right)\sum _{r=0}^n\left[{\left(-1\right)}^r\left({}^{n}C_r\right)\right]+\left(5n\right)\sum _{r=0}^n\left[{\left(-1\right)}^r\left({}^{n-1}C_{r-1}\right)\right]$.

$$\begin{gathered} \Rightarrow S=\left(3\times 0\right)+\left(5n\times 0\right) \\ \Rightarrow S=0 \end{gathered}$$

Therefore, the value of sum of the series $3·{}^{n}C_0-8·{}^{n}C_1+13·{}^{n}C_2-18·{}^{n}C_3+......$up to $\left(n+1\right)$ terms is $0$.

Hence, the correct option is (A).