Question

Evaluate the series $\sum _{n=0}^{\infty }{2}^{4-3n}$.

• A. $\frac{128}{7}$
• B. $16$
• C. $\frac{8}{7}$
• D. $0$
• E. The series diverges.

Answer 1

The correct option is A $\frac{128}{7}$

${\sum }_{n=0}^{\infty }{2}^{4-3n}=\stackrel{lim}{c\to \infty }{\int }_0^c\left(2^{4-3x}\right)dx=\stackrel{lim}{c\to \infty }{\int }_0^c{2}^4\times 2^{-3x}dx=\stackrel{lim}{c\to \infty }{2}^4{\int }_0^c{2}^{-3x}dx=\stackrel{lim}{c\to \infty }\frac{2^4(2^{-3c}-1)}{3\left(\log 2\right)}=\frac{64}{3\log 2}$

Hence the correct answer is $\frac{128}{7}$