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8. Sequences and Series
hard
જો $\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^{9}}+\ldots . .+\frac{10240}{3}=2^{ n } \cdot m$, કે જ્યાં $m$ એ અયુગ્મ છે તો $m . n$ ની કિમંત મેળવો.
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
