Dr - 2r-1 2 3r-1 4 5r-1 x y z 2n-1 3n -1 5n-1 - r-1nDr-

The correct option is D 0 Given, #160; D_ r = 2^ r 1 amp; 2( 3^ r 1 ) #160; amp; 4( 5^ r 1 ) #160; x amp; y amp; z 2^ n ...

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Byju's Answer

Standard XII

Mathematics

Summation of Determinant

Dr = 2r-1 ...

Question

$D_{r}$ = $∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & 2 (3^{r - 1}) & 4 (5^{r - 1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ $\Rightarrow \sum_{r = 1}^{n} D_{r} =$

2^{n} {.3}^{n .} 5^{n}

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(2^{n} - 1) (3^{n} - 1) (5^{n} - 1)

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x \times y \times z

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0

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Solution

The correct option is D $0$
Given, $D_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & 2 (3^{r - 1}) & 4 (5^{r - 1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow \sum_{r = 1}^{n} D_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{0} & 2 (3^{0}) & 4 (5^{0}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{1} & 2 (3^{1}) & 4 (5^{1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{2} & 2 (3^{2}) & 4 (5^{2}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + . . . . . . . .$

........ $+ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{n - 1} & 2 (3^{n - 1}) & 4 (5^{n - 1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{0} + 2^{1} + 2^{2} + . . . . + 2^{n - 1} & 2 (3^{0} + 3^{1} + 3^{2} + . . . . + 3^{n - 1}) & 4 (5^{0} + 5^{1} + 5^{2} + . . . . + 5^{n - 1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{n} - 1 & (3^{n} - 1) & (5^{n} - 1) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= 0$

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Similar questions

Q. If

D_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & 2 (3^{r - 1}) & 4 (5^{r - 1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

then prove that

n \sum r = 1 D_{r} = 0

Q. If

D_{r} =

∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & 2 (3^{r - 1}) & 4 (5^{r - 1}) x & y & 3 z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

then prove that

n \sum r = 1 D_{r} = 0.

Q. Let

D_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & {2.3}^{r - 1} & {4.5}^{r - 1} α & β & γ 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

. Then, the value of

\sum_{r = 1}^{n} D_{r}

$If D_{r} = ⎛ ⎜ ⎝ \begin{matrix} 2^{r - 1} & 2 . 3^{r - 1} & 4 . 5^{r - 1} α & β & γ 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣, then the value of \sum_{r = 1}^{n} D_{r} is$

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Dr = ∣∣ ∣ ∣∣2r−12(3r−1)4(5r−1)xyz2n−13n−15n−1∣∣ ∣ ∣∣ ⇒∑nr=1Dr=

$D_{r}$ = $∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{r - 1} & 2 (3^{r - 1}) & 4 (5^{r - 1}) x & y & z 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ $\Rightarrow \sum_{r = 1}^{n} D_{r} =$