8. Sequences and Series
hard

यदि $\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^9}+\ldots . .+\frac{10240}{3}=2^{ n } \cdot m$ है, जहाँ $m$ एक विषम संख्या है, तो $m . n$ बराबर है $...............$

A

$15$

B

$14$

C

$13$

D

$12$

(JEE MAIN-2022)

Solution

$\frac{6}{3^{12}}+10\left(\frac{1}{3^{11}}+\frac{2}{3^{10}}+\frac{2^{2}}{3^{9}}+\frac{2^{3}}{3^{8}}+\ldots .+\frac{2^{10}}{3}\right)$

$\frac{6}{3^{12}}+\frac{10}{3^{11}}\left(\frac{6^{11}-1}{6-1}\right)$

$=2^{12} \cdot 1 ; m . n =12$

Standard 11
Mathematics

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