Question

The value of $\sum_{r = 1}^{10}r.^{r}p_{r}$ is

• A. $11P_{11}$
• B. $11P_{11} - 1$
• C. $11P_{11} + 1$
• D. None

Answer 1

Answer: (B)

$11P_{11} - 1$

Answer 2

\begin{matrix} r \end{matrix} \end{matrix} \Rightarrow r.^{r}P_{r} = r.\begin{matrix} \begin{matrix} r \end{matrix} \end{matrix}$$ **=** $(r + 1 - 1)\begin{matrix} \begin{matrix} r \end{matrix} \end{matrix} = (r + 1)\begin{matrix} \begin{matrix} r \end{matrix} \end{matrix} - \begin{matrix} \begin{matrix} r \end{matrix} \end{matrix}$ **=** $\begin{matrix} \begin{matrix} r + 1 \end{matrix} \end{matrix} - \begin{matrix} \begin{matrix} r \end{matrix} \end{matrix}$ $$\sum_{r = 1}^{10}{r.^{r}P_{r}} = \sum_{r = 1}^{10}\begin{matrix} r + 1 \end{matrix} - \begin{matrix} r \end{matrix}$$ = $\begin{matrix} 11 \end{matrix} - \begin{matrix} 1 \end{matrix}$ = $11P_{11} - 1$