If Dr - - 2r-1 23r-1 45r-1- x y 3 z- 2n-1 3n-1 5n-1- - - then prove that -r-1n D r -0-

∑_r=1^n 2^r 1=1+2+2^2+2^3+ .............2^n 1 = 1.(2^n 1)/2 1=2^n 1, ∑ 2(3^r 1)= 2(1+3+3^2+ ..........+3^n 1) =2.1.(3^n 1)/3 1=3^n 1 , ∑ ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

If Dr = |[...

Question

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

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Solution

$n \sum r = 1 2^{r - 1} = 1 + 2 + 2^{2} + 2^{3} + . . . . . . . . . . . . {.2}^{n - 1}$
= $1. \frac{(2^{n} - 1)}{2 - 1} = 2^{n} - 1,$
$\sum 2 (3^{r - 1}) = 2 (1 + 3 + 3^{2} + . . . . . . . . . . + 3^{n - 1})$
$= 2.1. \frac{(3^{n} - 1)}{3 - 1} = 3^{n} - 1$ ,
$\sum 4 (5^{r - 1}) = 4 (1 + 5 + 5^{2} + . . . . . . + 5^{n - 1})$
$= 4.1 \frac{(5^{n} - 1)}{5 - 1} = 5^{n} - 1$ .
We have used the formula for G.P.
$S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ (r >1)
Rest is as in Q.16

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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

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} =

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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If Dr=∣∣ ∣ ∣∣2r−12(3r−1)4(5r−1)xy3z2n−13n−15n−1∣∣ ∣ ∣∣ then prove that n∑r=1Dr=0.

n∑r=12r−1=1+2+22+23+.............2n−1= 1.(2n−1)2−1=2n−1,∑2(3r−1)=2(1+3+32+..........+3n−1)=2.1.(3n−1)3−1=3n−1,∑4(5r−1)=4(1+5+52+......+5n−1)=4.1(5n−1)5−1=5n−1.We have used the formula for G.P.Sn=a(rn−1)r−1 (r >1)Rest is as in Q.16

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