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Q. The sum $5 \displaystyle\sum_{ n =1}^{\infty} \frac{2^{ n +2}}{4^{ n -2}}$ is equal to

Sequences and Series

Solution:

$T _{ n }=\frac{5 \cdot 2^{ n } \cdot 2^2}{4^{ n } \cdot 4^{-2}}=\left(\frac{1}{2}\right)^{ n } 5 \cdot 16 \cdot 4 ; \quad \therefore S =320 \displaystyle\sum_{ n =1}^{\infty}\left(\frac{1}{2}\right)^{ n }=320$