Question

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

• A. 1372
• B. 440
• C. 320
• D. 388

Answer 1

Answer: (C)

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