AIEEE 2002 (Offline)Question 1
The value of $2^{1/4}\cdot4^{1/8}\cdot8^{1/16}\cdots\infty$ is
Sequence and Series · Mixed Progressions (AP-GP)
Solution
Formulas used. (1) $2^p\cdot2^q\cdots=2^{p+q+\cdots}$(2) $\sum_{r\ge1}\frac{r}{2^r}=2$Step 1 - Powers of $2$. $2^{1/4}$, $4^{1/8}=2^{2/8}$, $8^{1/16}=2^{3/16}$, ...: the $r$-th factor is $2^{r/2^{r+1}}$. Step 2 - Add the exponents. $$\sum_{r\ge1}\frac{r}{2^{r+1}}=\frac12\sum_{r\ge1}\frac{r}{2^r}=\frac12\times2=1 .$$ Step 3. The value is $2^1=2$. Answer: (B)