Question

If \[A_r = \begin{vmatrix}1 & r & 2^r \\ 2 & n & n^2 \\ n & \frac{n \left( n + 1 \right)}{2} & 2^{n + 1}\end{vmatrix}\] , then the value of \[\sum^n_{r = 1} A_r\] is

पर्याय

• n

• 2n

• − 2n

•  n²

• None of these

Answer 1

\[A_r = \begin{vmatrix} 1 & r &2^r \\2 & n & n^2 \\n &  \frac{n\left( n + 1 \right)}{2} & 2^{n + 1} \end{vmatrix}\]
\[ \Rightarrow \sum^n_{r = 1} A_r = \begin{vmatrix} \sum^n_{r = 1} 1 & \sum^n_{r = 1} r & \sum^n_{r = 1} 2^r \\ \sum^n_{r = 1} 2 & n & n^2 \\n & \frac{n\left( n + 1 \right)}{2} & 2^{n + 1} \end{vmatrix}\]
\[As \sum^n_{r = 1} 1 = 1 + 1 + 1 . . . + 1 (\text{n times}) = n\]
\[ \sum^n_{r = 1} r = 1 + 2 + 3 + . . . + n = \frac{n\left( n + 1 \right)}{2} \]
\[Let S = \sum^n_{r = 1} 2^r = 2 + 2^2 + 2^3 = . . . + 2^n \]
\[2S = 2^2 + 2^3 = . . . + 2^n + 2^{n + 1} \]
\[ \Rightarrow 2S - S = S = \sum^n_{r = 1} 2^r = 2^{n + 1} - 2\]
\[ \Rightarrow \sum^n_{r = 1} A_r = \begin{vmatrix} n & \frac{n\left( n + 1 \right)}{2} & 2^{n + 1} - 2\\2n & n & n^2 \\ n & \frac{n\left( n + 1 \right)}{2} & 2^{n + 1} \end{vmatrix} \]
\[\left[\text{ Applying }R_1 \to R_1 - R_2 \right]\]
\[ \sum^n_{r = 1} A_r = \begin{vmatrix} n - n & \frac{n\left( n + 1 \right)}{2} - \frac{n\left( n + 1 \right)}{2} & 2^{n + 1} - 2 - 2^{n + 1} \\2n & n & n^2 \\ n & \frac{n\left( n + 1 \right)}{2} & 2^{n + 1} \end{vmatrix}\]
\[ = \begin{vmatrix} 0 & 0 & - 2\\2n & n & n^2 \\ n & \frac{n\left( n + 1 \right)}{2} & 2^{n + 1} \end{vmatrix}\]
\[ = - 2 \times \begin{vmatrix} 2n & n \\ n & \frac{n\left( n + 1 \right)}{2} \end{vmatrix}\]
\[ = - 2\left[ n^3 + n^2 - n^2 \right]\]
\[ = - 2 n^3 \]