જો {D_r} = left| {begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}x&y&z{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}end{array}} right|, તો sumlimits_{r = 1}^n {{D

Q: જો ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|$, તો $\sum\limits_{r = 1}^n {{D_r} = } $

c (c) ${D_r} = \left| {\,\begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}\,} \right|$ $ \Rightarrow \sum\limits_{r = 1}^n {{D_r} = \left| {\,\begin{array}{*{20}{c}}{\sum\limits_{r = 1}^n {{2^{r - 1}}} }&{\sum\limits_{r = 1}^n {{{2.3}^{r - 1}}} }&{\sum\limits_{r = 1}^n {{{4.5}^{r - 1}}} }\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|} $ $ \Rightarrow \sum\limits_{r = 1}^n {{D_r} = } \left| {\,\begin{array}{*{20}{c}}{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|$ Since we know that $\sum\limits_{r = 1}^n {{2^{r - 1}} = \frac{{{2^n} - 1}}{{2 - 1}} = {2^n} - 1,} $ $2\sum\limits_{r = 1}^n {{3^{r - 1}} = 2\frac{{{3^n} - 1}}{{3 - 1}} = {3^n} - 1} $ and $4\sum\limits_{r = 1}^n {{5^{r - 1}} = 4\frac{{{5^n} - 1}}{{5 - 1}} = {5^n} - 1} $ $ \Rightarrow \,\,\sum\limits_{r = 1}^n {{D_r} = 0} $, $(\because {R_1} \equiv {R_3})$ .

MCQ

ગુજરાતી માધ્યમ

જો ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|$, તો $\sum\limits_{r = 1}^n {{D_r} = } $

A$1$
B$-1$
C$0$
D
એકપણ નહી.

Diffcult

ધોરણ 12 સાયન્સ
ગણિત
3 and 4 . determinant and metrices
JEE

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