Question

If \[\dfrac{1}{{{1^4}}} + \dfrac{1}{{{2^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{4^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{{\pi ^4}}}{{90}}\] , then find the value of \[\dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty \] .
A. \[\dfrac{{{\pi ^4}}}{{96}}\]
B. \[\dfrac{{{\pi ^4}}}{{45}}\]
C. \[\dfrac{{89{\pi ^4}}}{{90}}\]
D. None of these

Answer 1

Hint:First we will separate the terms such that the denominator is 4 rise of odd numbers and even numbers. Then simplify the 4 rises of even numbers and calculate the value of the given expression.

Complete step by step solution:
Given equation
\[\dfrac{1}{{{1^4}}} + \dfrac{1}{{{2^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{4^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{{\pi ^4}}}{{90}}\] ……………(1)
Simplifying the given equation by separating even and odd numbers and we get
\[ \Rightarrow \left( {\dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty } \right) + \left( {\dfrac{1}{{{2^4}}} + \dfrac{1}{{{4^4}}} + \dfrac{1}{{{6^4}}} + ...\infty } \right) = \dfrac{{{\pi ^4}}}{{90}}\]
\[ \Rightarrow \left( {\dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty } \right) + \left( {\dfrac{1}{{{2^4}}} + \dfrac{1}{{{{\left( {2 \times 2} \right)}^4}}} + \dfrac{1}{{{{\left( {3 \times 2} \right)}^4}}} + ...\infty } \right) = \dfrac{{{\pi ^4}}}{{90}}\]
Taking common \[\dfrac{1}{{{2^4}}}\] from even numbers and we get
\[ \Rightarrow \left( {\dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty } \right) + \left( {\dfrac{1}{{{2^4}}}} \right)\left( {\dfrac{1}{{{1^4}}} + \dfrac{1}{{{2^4}}} + \dfrac{1}{{{3^4}}} + ...\infty } \right) = \dfrac{{{\pi ^4}}}{{90}}\]……………..(2)
Substituting the value of the second term of equation (2) from equation (1), we get
\[ \Rightarrow \left( {\dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty } \right) + \left( {\dfrac{1}{{{2^4}}}} \right)\left( {\dfrac{{{\pi ^4}}}{{90}}} \right) = \dfrac{{{\pi ^4}}}{{90}}\]
Simplifying the expression and we get
\[ \Rightarrow \left( {\dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty } \right) + \left( {\dfrac{{{\pi ^4}}}{{{2^4} \times 90}}} \right) = \dfrac{{{\pi ^4}}}{{90}}\]
\[ \Rightarrow \dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{{\pi ^4}}}{{90}} - \dfrac{{{\pi ^4}}}{{{2^4} \times 90}}\]
\[ \Rightarrow \dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{{2^4}{\pi ^4} - {\pi ^4}}}{{{2^4} \times 90}}\]
\[ \Rightarrow \dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{\left( {{2^4} - 1} \right){\pi ^4}}}{{{2^4} \times 90}}\]
\[ \Rightarrow \dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{\left( {16 - 1} \right){\pi ^4}}}{{16 \times 90}}\]
\[ \Rightarrow \dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{15 \times {\pi ^4}}}{{16 \times 90}}\]
\[ \Rightarrow \dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{{\pi ^4}}}{{16 \times 6}}\]
\[ \Rightarrow \dfrac{1}{{{1^4}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{5^4}}} + ...\infty = \dfrac{{{\pi ^4}}}{{96}}\]
Hence, the correct option is option A.

Additional information:
Definition of infinite series: The total of infinitely many numbers connected in a specific way and listed in a specific order is known as an infinite series. In mathematics as well as in fields like physics, chemistry, and biology, infinite series is helpful.

Note:To solve infinite series you need to use the substitution method. In this question, you need to write the equation as a sum of a fraction whose denominator is 4 rises of odd numbers and whose denominator is 4 rises. Then take out the common factor from the sum whose denominator is 4 rises of even numbers and substitute the value of the given expression. After simplifying we can get the required value.