Question

If

$\frac{1}{ {1}^{4} } + \frac{1}{ {2}^{4} } + \frac{1}{ {3}^{4} } + - - - \infty = \frac{ {\pi}^{4} }{90}$

then find the value of

$\frac{1}{ {1}^{4} } + \frac{1}{ {3}^{4} } + \frac{1}{ {5}^{4} } + - - - \infty$
​

Answer 1

$\red{\large\underline{\sf{Given- }}}$

$\rm \: \dfrac{1}{ {1}^{4} } +\dfrac{1}{ {2}^{4} } + \dfrac{1}{ {3}^{4} } + \dfrac{1}{ {4}^{4} } + \dfrac{1}{ {5}^{4} } + - - - \infty = \dfrac{ {\pi}^{4} }{90}$

$\purple{\large\underline{\sf{To\:Find - }}}$

$\rm \: \dfrac{1}{ {1}^{4} } + \dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - - \infty$

$\green{\large\underline{\sf{Solution-}}}$

Given series is

$\rm \: \dfrac{1}{ {1}^{4} } +\dfrac{1}{ {2}^{4} } + \dfrac{1}{ {3}^{4} } + \dfrac{1}{ {4}^{4} } + \dfrac{1}{ {5}^{4} } + - - - \infty = \dfrac{ {\pi}^{4} }{90}$

can also be re-arranged as

$\rm \:\bigg[ \dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty \bigg] +\bigg[ \dfrac{1}{ {2}^{4} } + \dfrac{1}{ {4}^{4} } + - - - \infty\bigg] = \dfrac{ {\pi}^{4} }{90}$

$\rm \:\bigg[ \dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty \bigg] +\bigg[ \dfrac{1}{ {(2.1)}^{4} } + \dfrac{1}{ {(2.2)}^{4} } + - - - \infty\bigg] = \dfrac{ {\pi}^{4} }{90}$

$\rm \:\bigg[ \dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty \bigg] +\bigg[ \dfrac{1}{ {1}^{4}{2}^{4} } + \dfrac{1}{ {2}^{4} {2}^{4} } + - - - \infty\bigg] = \dfrac{ {\pi}^{4} }{90}$

$\rm \:\bigg[ \dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty \bigg] +\dfrac{1}{ {2}^{4} } \bigg[ \dfrac{1}{{1}^{4} } + \dfrac{1}{{2}^{4} } + - - - \infty\bigg] = \dfrac{ {\pi}^{4} }{90}$

Using given series, we get

$\rm \:\dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty +\dfrac{1}{ {2}^{4} } \times \dfrac{ {\pi}^{4} }{90}= \dfrac{ {\pi}^{4} }{90}$

$\rm \:\dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty= \dfrac{ {\pi}^{4} }{90} - \dfrac{1}{ {2}^{4} } \times \dfrac{ {\pi}^{4} }{90}$

$\rm \:\dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty= \dfrac{ {\pi}^{4} }{90} \bigg[1- \dfrac{1}{ {2}^{4} }\bigg]$

$\rm \:\dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty= \dfrac{ {\pi}^{4} }{90} \bigg[1- \dfrac{1}{16}\bigg]$

$\rm \:\dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty= \dfrac{ {\pi}^{4} }{90} \bigg[ \dfrac{16 - 1}{16}\bigg]$

$\rm \:\dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty= \dfrac{ {\pi}^{4} }{90} \bigg[ \dfrac{15}{16}\bigg]$

$\rm\implies \:\boxed{\tt{ \rm \:\dfrac{1}{ {1}^{4} } +\dfrac{1}{ {3}^{4} } + \dfrac{1}{ {5}^{4} } + - - \infty= \dfrac{ {\pi}^{4} }{96} \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \rm \: 1 + 2 + 3 + - - - + n = \frac{n(n + 1)}{2}}}$

$\boxed{\tt{ \rm \: {1}^{2} + {2}^{2} + {3}^{2} + - - + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6}}}$

$\boxed{\tt{ \rm \: {1}^{3} + {2}^{3} + {3}^{3} + - - + {n}^{3} = \bigg[\frac{n(n + 1)}{2}\bigg]^{2} }}$

Answer 2

Answer:

π⁴/96

Step-by-step explanation:

Given that 1/1⁴ + 1/2⁴ + 1/3⁴ + - - - - ∞ = π⁴/90

We need to find out 1⁴ + 1/3⁴ + 1/5⁴ + - - - - ∞

Let's say 1/1⁴ + 1/3⁴ + 1/5⁴ + - - - - - ∞ = A

Now,

1/1⁴ + 1/2⁴ + 1/3⁴ + - - - - ∞ = π⁴/90

(1/1⁴ + 1/3⁴ + 1/5⁴ + - - - - ) + (1/2⁴ + 1/4⁴ + 1/6⁴ + - - - -) = π⁴/90

Take 1/2⁴ as common,

A + 1/2⁴ (1/1⁴ + 1/2⁴ + 1/3⁴ + - - - - - ∞) = π⁴/90

A + 1/16 × π⁴/90 = π⁴/90

A = π⁴/90 - 1/16 × π⁴/90

Take π⁴/90 as common,

A = π⁴/90 × (1 - 1/16)

A = π⁴/90 × [(16 - 1)/16]

A = π⁴/90 × 15/16

A = π⁴/96

Hence, the required answer is π⁴/96