Question

What is the value of \[{2^{\dfrac{1}{4}}} \times {4^{\dfrac{1}{8}}} \times {8^{\dfrac{1}{{16}}}} \times ......\infty \]?
A.1
B.2
C.\[\dfrac{3}{2}\]
D.4

Answer 1

Hint: First we will apply the exponential formula \[{a^m} \times {a^n} = {a^{m + n}}\] to simplify the given expression. Now we need to solve the series on the power of \[2\]. Name the series as \[S\]. Then we will calculate \[S - \dfrac{S}{2}\] and the series becomes a GP. Then apply sum the formula of infinite series of a GP that is \[\dfrac{a}{{1 - r}}\] . Then we will calculate \[{2^S}\] to the value of the given expression.

Formula Used:
We will use the exponential formula \[{a^m} \times {a^n} = {a^{m + n}}\], and then
Sum of infinite G.P which is given by \[\dfrac{a}{{1 - r}}\], where \[r\] is common ratio, \[a\] is the first term.

Complete step by step solution:
Given \[{2^{\dfrac{1}{4}}} \times {4^{\dfrac{1}{8}}} \times {8^{\dfrac{1}{{16}}}} \times ......\infty \]
Now we will make each term to same base, i.e.,
\[ \Rightarrow {2^{\dfrac{1}{4}}} \times {\left( {{2^2}} \right)^{\dfrac{1}{8}}} \times {\left( {{2^3}} \right)^{\dfrac{1}{{16}}}} \times ......\infty \]
Now we will simplify, then we will get,
\[ \Rightarrow {2^{\dfrac{1}{4}}} \times {2^{\dfrac{2}{8}}} \times {2^{\dfrac{3}{{16}}}} \times {2^{\dfrac{4}{{32}}}}......\infty \]
Now we will use the exponential formula i.e., \[{a^m} \times {a^n} = {a^{m + n}}\], we will get,
\[ \Rightarrow {2^{\dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{{16}} + \dfrac{4}{{32}} + ..........\infty }}\]
Now we will assume the expression in the power as a variable say ‘S’
\[ \Rightarrow S = \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{{16}} + \dfrac{4}{{32}} + ...... + \infty - - - - (1)\]
Now divide both sides of the equation with 2, then we will get,
\[ \Rightarrow \dfrac{S}{2} = \dfrac{1}{8} + \dfrac{2}{{16}} + \dfrac{3}{{32}} + \dfrac{4}{{64}} + ...... + \infty - - - - (2)\]
Now we will subtract (2) from (2), we will get,
\[ \Rightarrow S - \dfrac{S}{2} = \dfrac{1}{4} - \dfrac{1}{8} + \dfrac{2}{8} - \dfrac{2}{{16}} + \dfrac{3}{{16}} - \dfrac{3}{{32}} + \dfrac{4}{{32}} - \dfrac{4}{{64}} + ...... + \infty \]
Now we will simplify the expression we will get,
\[\dfrac{S}{2} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \dfrac{1}{{32}} + ...... + \infty \]
Now we observe that the series is in infinite G.P with first term\[\dfrac{1}{4}\], and common ratio=\[\dfrac{{\dfrac{1}{8}}}{{\dfrac{1}{4}}} = \dfrac{1}{2}\], substituting the values in the infinite G.P formula i.e., \[\dfrac{a}{{1 - r}}\], we will get,
\[ \Rightarrow \dfrac{S}{2} = \dfrac{{\dfrac{1}{4}}}{{1 - \dfrac{1}{2}}}\]
Now we will simplify, then we will get,
\[ \Rightarrow \dfrac{S}{2} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{2}}}\]
Now we will further simplify we will get,
\[ \Rightarrow \dfrac{S}{2} = \dfrac{1}{2}\]
Now we will further simplify for\[S\],we will get,
\[ \Rightarrow S = 1\],
Now substitute the value of \[S\]in the expression,
\[ \Rightarrow {2^{\dfrac{1}{4}}} \times {4^{\dfrac{1}{8}}} \times {8^{\dfrac{1}{{16}}}} \times ......\infty = {2^S}\]
Now we will get,
\[ \Rightarrow {2^{\dfrac{1}{4}}} \times {4^{\dfrac{1}{8}}} \times {8^{\dfrac{1}{{16}}}} \times ......\infty = {2^1}\]
Now we will further simplify we will get,
\[ \Rightarrow {2^{\dfrac{1}{4}}} \times {4^{\dfrac{1}{8}}} \times {8^{\dfrac{1}{{16}}}} \times ......\infty = 2\]

The correct option is B.

Note: The second and first term of the series \[\dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{{16}} + \dfrac{4}{{32}} + ...... + \infty \] are same. So students assume that the series is \[\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + ...... + \infty \] and take common ratio \[1\]. Then they apply the formula \[\dfrac{a}{{1 - r}}\] and reach a wrong answer.