Question

Find the sum of \[{2^{1/4}}{.4^{1/8}}{.8^{1/16}}{.16^{1/32}}.....\]

A. \[1\]
B. \[2\]
C. \[\dfrac{3}{2}\]
D. \[\dfrac{5}{2}\]

Answer 1

Hint: In this question, we need to find the total sum of the given infinite sequence. For this, we need to use the concept of geometric progression.

Formula Used: The sum of infinite that means the sum of a geometric sequence with infinite terms is \[{S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}};-1 < r < 1\] where, \[r\] is common ratio.

Complete step-by-step answer: We know that, the given sequence is \[{2^{1/4}}{.4^{1/8}}{.8^{1/16}}{.16^{1/32}}.....\]
Let us simplify this further.
\[
  {S_\infty } = {2^{1/4}}{.4^{1/8}}{.8^{1/16}}{.16^{1/32}}..... \\
   \Rightarrow {S_\infty } = {2^{1/4}}.{\left( {{{\left( 2 \right)}^2}} \right)^{1/8}}.{\left( {{{\left( 2 \right)}^3}} \right)^{1/16}}.{\left( {{{\left( 2 \right)}^4}} \right)^{1/32}}..... \\
   \Rightarrow {S_\infty } = {2^{1/4}}.\left( {{{\left( 2 \right)}^{2/8}}} \right).\left( {{{\left( 2 \right)}^{3/16}}} \right).\left( {{{\left( 2 \right)}^{4/32}}} \right)..... \\
   \Rightarrow {S_\infty } = {2^{\left( {1/4} \right) + \left( {2/8} \right) + \left( {3/16} \right) + \left( {4/32} \right)}}... \\
 \]
Let \[S = {2^E}\]
Thus, we get
Let \[E = \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{{16}} + \dfrac{4}{{32}} + ....\]
\[E = \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{{16}} + \dfrac{4}{{32}} + ....\left( 1 \right)\]
Multiply by \[\dfrac{1}{2}\]to both sides.
\[\dfrac{E}{2} = \dfrac{1}{8} + \dfrac{2}{{16}} + \dfrac{3}{{32}} + \dfrac{4}{{64}} + ....\left( 2 \right)\]
Now, subtract equation (2) from equation (1)
So, we get
\[
  E - \dfrac{E}{2} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \dfrac{1}{{32}} + ... \\
   \Rightarrow \dfrac{E}{2} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \dfrac{1}{{32}} + ... \\
 \]
By simplifying it further, we get
\[ \Rightarrow \dfrac{E}{2} = \dfrac{1}{2}\left( {\dfrac{1}{2}} \right) + \dfrac{1}{2}\left( {\dfrac{1}{4}} \right) + \dfrac{1}{2}\left( {\dfrac{1}{8}} \right) + \dfrac{1}{2}\left( {\dfrac{1}{{16}}} \right) + ...\]
Hence, the common ratio \[r\] is \[r = \dfrac{1}{2}\]
Also, the first term of a sequence is \[a = \dfrac{1}{4}\]
By applying the formula of geometric progression, we get
\[
   \Rightarrow \dfrac{E}{2} = \left( {\dfrac{{1/4}}{{1 - \left( {1/2} \right)}}} \right) \\
   \Rightarrow \dfrac{E}{2} = \dfrac{1}{4} \times \dfrac{2}{1} \\
   \Rightarrow E = 1 \\
 \]
Now, put the value of \[E\] in the equation \[S = {2^E}\]
Thus, we get
\[
  S = {2^E} \\
   \Rightarrow S = {2^1} \\
   \Rightarrow S = 2 \\
 \]
Hence, the sum of the sequence \[{2^{1/4}}{.4^{1/8}}{.8^{1/16}}{.16^{1/32}}.....\]is \[2\].


Therefore, the correct option is (B)

Note: Many students make mistakes in the calculation part of the geometric progression sequence. This is the only way through which we can solve this example in an easy manner.