Question

If $\mathrm{p}+\mathrm{q}=1$ then $\mathrm{\Sigma }\mathrm{r}.{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}{\mathrm{p}}^{\mathrm{r}}{\mathrm{q}}^{\mathrm{n}-\mathrm{r}},\mathrm{r}\in [\mathrm{r}=0,\mathrm{n}]$ is equal to

• A. $1$
• B. $\mathrm{np}$
• C. $\mathrm{npq}$
• D. $0$

Answer 1

The correct option is B

$\mathrm{np}$

Explanation for Correct Option :

Step 1:Solving the given equation using the binomial expansion

${}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!\mathrm{r}!}$ when $n>r$

$\begin{array}{rcl}\underset{\mathrm{r}=0}{\overset{\mathrm{n}}{\therefore \sum }}\mathrm{r}{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}{p}^{\mathrm{r}}{q}^{\mathrm{n}-\mathrm{r}}& =& {}^{\mathrm{n}}\mathrm{C}_1\mathrm{p}{\mathrm{q}}^{\mathrm{n}-1}+2\times {}^{\mathrm{n}}\mathrm{C}_2{\mathrm{p}}^2{\mathrm{q}}^{\mathrm{n}-2}+......n\times {}^{\mathrm{n}}\mathrm{C}_{\mathrm{n}}{\mathrm{p}}^{\mathrm{n}}{\mathrm{q}}^{\mathrm{n}-\mathrm{n}}\\ & =& {\mathrm{npq}}^{\mathrm{n}-1}+\mathrm{n}(\mathrm{n}-1){\mathrm{p}}^2{\mathrm{q}}^{\mathrm{n}-2}+......\\ & =& \mathrm{np}\left({\mathrm{q}}^{\mathrm{n}-1}+(\mathrm{n}-1){\mathrm{pq}}^{\mathrm{n}-2}+......\right)......\left(1\right)\end{array}$

Step 2: Put the value $\mathrm{p}+\mathrm{q}=1$ in equation (1)

$$\begin{gathered} \Rightarrow \mathrm{np}{(\mathrm{q}+\mathrm{p})}^{\mathrm{n}-1} \\ =\mathrm{np}{\left(1\right)}^{\mathrm{n}-1} \\ =\mathrm{np} \end{gathered}$$

Hence Option (B) is correct.