If $D_{r} = \left| \begin{matrix} r & 1 & \frac{n(n + 1)}{2} \ 2r - 1 & 4 & n^{2} \ 2^{r - 1} & 5 & 2^{n} - 1 \end{matrix} ight|,\$then the value of $\sum_{r = 0}^{n}D_{r}\ is$

Question

If $D_{r} = \left| \begin{matrix} r & 1 & \frac{n(n + 1)}{2} \ 2r - 1 & 4 & n^{2} \ 2^{r - 1} & 5 & 2^{n} - 1 \end{matrix} ight|,\$then the value of $\sum_{r = 0}^{n}D_{r}\ is$

Dr. Abubacker · Accepted Answer

(a)
$\sum_{r = 0}^{n}{D_{r} = \left| \begin{matrix}
\sum_{}^{}r & 1 & \frac{n(n + 1)}{2} \
2\sum_{}^{}r - \sum_{}^{}1 & 4 & n^{2} \
\sum_{}^{}2^{r - 1} & 5 & 2^{n} - 1
\end{matrix} ight|}$
$= \left| \begin{matrix}
\frac{n(n + 1)}{2} & 1 & \frac{n(n + 1)}{2} \
n^{2} & 4 & n^{2} \
2^{n} - 1 & 5 & 2^{n} - 1
\end{matrix} ight| = 0$
[$\because$ two columns are identical]

Q. If $D_{r} = \left| \begin{matrix} r & 1 & \frac{n(n + 1)}{2} \\ 2r - 1 & 4 & n^{2} \\ 2^{r - 1} & 5 & 2^{n} - 1 \end{matrix} \right|,\$then the value of $\sum_{r = 0}^{n}D_{r}\ is$

Explanation

Reviewed by Dr. Abubacker

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