If (delta)_r = |1  r  2^r |

|2  n  n² |

|n  n(n+1)/2  2ⁿ⁺¹ |

Find, summation  ​  (delta)r
        r = 1 to n
 

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} {△}_{\mathrm{r}}=\left|\begin{array}{ccc}1& \mathrm{r}& 2^{\mathrm{r}}\\ 2& \mathrm{n}& {\mathrm{n}}^2\\ \mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}\end{array}\right| \\ \left[\mathrm{Check} \mathrm{if} \mathrm{last} \mathrm{entry} \mathrm{is}{2}^{\mathrm{n}+1} \mathrm{or} 2^{\mathrm{n}+1}-2, \mathrm{I} \mathrm{will} \mathrm{solve} \mathrm{it} \mathrm{using} 2^{\mathrm{n}+1}\right] \\ \Rightarrow \sum _{\mathrm{r}=1}^{\mathrm{n}}{△}_{\mathrm{r}}=\left|\begin{array}{ccc}\sum _{\mathrm{r}=1}^{\mathrm{n}} 1& \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}& \sum _{\mathrm{r}=1}^{\mathrm{n}} 2^{\mathrm{r}}\\ 2& \mathrm{n}& {\mathrm{n}}^2\\ \mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}\end{array}\right|....\left(\mathrm{i}\right) \\ \mathrm{Now} \\ \sum _{\mathrm{r}=1}^{\mathrm{n}} 1=1+1+1+.... \mathrm{n} \mathrm{times}=\mathrm{n} \\ \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}=1+2+3+.....+\mathrm{n}=\frac{\mathrm{n}}{2}\left(\mathrm{First} \mathrm{term}+\mathrm{Last} \mathrm{term}\right) \left[\mathrm{It} \mathrm{is} \mathrm{an} \mathrm{A}.\mathrm{P}.\right] \\ \Rightarrow \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2} \\ \sum _{\mathrm{r}=1}^{\mathrm{n}} 2^{\mathrm{r}}=2+2^2+2^3+....+2^{\mathrm{n}}=2.\left(\frac{2^{\mathrm{n}}-1}{2-1}\right) \left[\mathrm{It} \mathrm{is} \mathrm{a} \mathrm{G}.\mathrm{P}.\right] \\ \Rightarrow \sum _{\mathrm{r}=1}^{\mathrm{n}} 2^{\mathrm{r}}=2^{\mathrm{n}+1}-2 \\ \mathrm{Hence} \left(\mathrm{i}\right) \mathrm{becomes} \\ \Rightarrow \sum _{\mathrm{r}=1}^{\mathrm{n}}{△}_{\mathrm{r}}=\left|\begin{array}{ccc}\mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}-2\\ 2& \mathrm{n}& {\mathrm{n}}^2\\ \mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}\end{array}\right| \\ =\left|\begin{array}{ccc}\mathrm{n}-0& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}-0& 2^{\mathrm{n}+1}-2\\ 2& \mathrm{n}& {\mathrm{n}}^2\\ \mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}\end{array}\right| \\ =\left|\begin{array}{ccc}\mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}\\ 2& \mathrm{n}& {\mathrm{n}}^2\\ \mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}\end{array}\right|+\left|\begin{array}{ccc}0& 0& -2\\ 2& \mathrm{n}& {\mathrm{n}}^2\\ \mathrm{n}& \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}& 2^{\mathrm{n}+1}\end{array}\right| \\ \mathrm{In} \mathrm{first} \mathrm{determinant}, \mathrm{first} \mathrm{and} \mathrm{third} \mathrm{row} \mathrm{are} \mathrm{same}, \mathrm{hence} \mathrm{it} \mathrm{will} \mathrm{become} \\ 0 \mathrm{and} \mathrm{expand} \mathrm{second} \mathrm{determinant} \mathrm{along} {\mathrm{R}}_1 \\ \Rightarrow \sum _{\mathrm{r}=1}^{\mathrm{n}}{△}_{\mathrm{r}}=0+\left[-2\left\{\mathrm{n}.2^{\mathrm{n}+1}-{\mathrm{n}}^2.\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}\right] \\ =-2\left\{\mathrm{n}.2^{\mathrm{n}+1}-{\mathrm{n}}^2.\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\} \\ ={\mathrm{n}}^2.\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}-\mathrm{n}.2^{\mathrm{n}+2} \left(\mathrm{Answer}\right) \end{gathered}$$

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