Sequence And Series Question 166
Question: If $ \frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+…..+\infty =\frac{{{\pi }^{4}}}{90} $ , then the value of $ \frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+…..\infty $ is
[AMU 2005]
Options:
A) $ \frac{{{\pi }^{4}}}{96} $
B) $ \frac{{{\pi }^{4}}}{45} $
C) $ \frac{89}{90}{{\pi }^{4}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+…..\infty $ $ =\frac{{{\pi }^{4}}}{90} $ $ \frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+…+\infty +\frac{1}{2^{4}}( \frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+…\infty ) $ $ =\frac{{{\pi }^{4}}}{90} $ $ \frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\frac{1}{7^{4}}+……+\infty $ $ +\frac{1}{16}\times \frac{{{\pi }^{4}}}{90}=\frac{{{\pi }^{4}}}{90} $ \ $ \frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\frac{1}{7^{4}}+…..+\infty $ $ =\frac{{{\pi }^{4}}}{90}-\frac{1}{16}( \frac{{{\pi }^{4}}}{90} ) $ $ =\frac{15}{16}( \frac{{{\pi }^{4}}}{90} )=\frac{{{\pi }^{4}}}{96} $ .
BETA
