Question

Find the value of $(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{16})(1+\frac{1}{256})...\infty$

• A. $1$
• B. $2$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$2$

$(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{16})(1+\frac{1}{256})\dots$

$=(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8})(1+\frac{1}{16})(1+\frac{1}{256})\dots$

$=(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128})(1+\frac{1}{256})\dots$

$=(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256})\dots$

$=\frac{1}{1-\frac{1}{2}}$ (sum of infinite GP)

$=2$