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  • Provide solution for RD Sharma maths class 12 chapter Determinants exercise multiple choise question 14

Provide solution for RD Sharma maths class 12 chapter Determinants exercise multiple choise question 14

Answers (1)

Answer:

Correct option (c)

Hint:

Simply solve this determinant

Given:

        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|

We have to find

         \sum_{r=1}^{n}A^{r}

Solution:

If

        A_{r}=\left|1\, r\, 2^{r}\, 2\, n\, n^{2}\, n\, \frac{n(n+1)}{2}\, 2^{n+1}\right|

        \Rightarrow \sum_{r=1}^{n}A_{r}=\left| n\frac{n(n+1)}{2} 2^{n+1}-2\; 2\; n\; n^{2}\; n\frac{n(n+1)}{2} 2^{n+1} \right|

Since,\; \; \sum_{r=1}^{n}r=1+2+3+.....+n\; \; =\frac{n(n+1)}{2}

        \sum_{r=1}^{n}2^{r}=2+2^{2}+2^{3}+.....+2^{n}\; \; =\frac{2(2^{n}-1)}{2-1}=2^{n+1}-2

Now applying, R1→R1-R3

        \Rightarrow \sum_{r=1}^{n}A_{r}=\left | 0-2\; 2\; n\; n^{2}\; n\frac{n(n+1)}{2}2^{n+1} \right |

        \Rightarrow \sum_{r=1}^{n}A_{r}=-2[n(n+1)-n^{2}]

        \Rightarrow \sum_{r=1}^{n}A_{r}=-2[n^{2}+n-n^{2}]

        \Rightarrow \sum_{r=1}^{n}A_{r}=-2n

Posted by

Gurleen Kaur

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