A) \[{{(30)}^{n}}\]
B) \[{{(10)}^{n}}\]
C) \[0\]
D) \[{{2}^{n}}+{{3}^{n}}+{{5}^{n}}\]
E) none of these
Correct Answer: C
Solution :
\[\because \]\[a=1+2+4+....\]up to n terms \[=1.\frac{{{2}^{n}}-1}{2-1}=({{2}^{n}}-1)\] \[b=1+3+9+\text{ }...\]upto n terms \[=1.\left( \frac{{{3}^{n}}-1}{3-1} \right)=\frac{{{3}^{n}}-1}{2}\] and \[c=1+\text{ }5+25+...\]up to n terms \[=1.\frac{{{5}^{n}}-1}{5-1}=\frac{{{5}^{n}}-1}{4}\] \[\therefore \,\,\left| \begin{matrix} a & 2b & 4c \\ 2 & 2 & 2 \\ {{2}^{n}} & {{3}^{n}} & {{5}^{n}} \\ \end{matrix} \right|=2\left| \begin{matrix} {{2}^{n}}-1 & {{3}^{n}}-1 & {{5}^{n}}-1 \\ 1 & 1 & 1 \\ {{2}^{n}} & {{3}^{n}} & {{5}^{n}} \\ \end{matrix} \right|\] \[=2\left| \begin{matrix} {{2}^{n}} & {{3}^{n}} & {{5}^{n}} \\ 1 & 1 & 1 \\ {{2}^{n}} & {{3}^{n}} & {{5}^{n}} \\ \end{matrix} \right|\,=({{R}_{1}}\to {{R}_{1}}+{{R}_{2}})\] \[=2\times 0=0\]
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