If $A_{r}=\left|\begin{array}{ccc}1 & r & 2^{r} \\ 2 & n & n^{2} \\ n & \frac{n(n+1)}{2} & 2^{n+1}\end{array}\right|$, then the value of $\sum_{r=1}^{n} A_{r}$ is
(a) $n$
(b) $2 n$
(c) $-2 n$
(d) $n^{2}$
$A_{r}=\mid 1 \quad r \quad 2^{r}$
$\begin{array}{lll}2 & n & n^{2}\end{array}$
$n \quad \frac{n(n+1)}{2} \quad 2^{n+1} \mid$
$\Rightarrow \sum_{r=1}^{n} A_{r}=\mid \sum_{r=1}^{n} 1 \quad \sum_{r=1}^{n} r \quad \sum_{r=1}^{n} 2^{r} \sum_{r=1}^{n} 2 \quad n \quad n \quad n^{2} n \quad \frac{n(n+1)}{2}$
As $\sum_{r=1}^{n} 1=1+1+1 \ldots+1(n$ times $)=n$
$\sum_{r=1}^{n} r=1+2+3+\ldots+n=\frac{n(n+1)}{2}$
Let $S=\sum_{r=1}^{n} 2^{r}=2+2^{2}+2^{3}=\ldots+2^{n}$
$2 S=2^{2}+2^{3}=\ldots+2^{n}+2^{n+1}$
$\Rightarrow 2 S-S=S=\sum_{r=1}^{n} 2^{r}=2^{n+1}-2$
$\Rightarrow \sum_{r=1}^{n} A_{r}=\mid n \quad \frac{n(n+1)}{2} \quad 2^{n+1}-2$
$2 n \quad n \quad n^{2}$
$n \quad \frac{n(n+1)}{2}$ $2^{n+1} \mid$
[Applying $R_{1} \rightarrow R_{1}-R_{2}$ ]
$\sum_{r=1}^{n} A_{r}$
$=\left|n-n \quad \frac{n(n+1)}{2}-\frac{n(n+1)}{2} \quad 2^{n+1}-2-2^{n+1} 2 n \quad n \quad n^{2} n \quad \frac{n(n+1)}{2} \quad 2^{n+1}\right|$
$=\mid \begin{array}{lll}0 & 0 & -2\end{array}$
$\begin{array}{lll}2 n & n & n^{2}\end{array}$
$n \quad \frac{n(n+1)}{2}$ $2^{n+1} \mid$
$=-2 \times \mid 2 n$ $n$
$=-2\left[n^{3}+n^{2}-n^{2}\right]$
$=-2 n^{3}$
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