A) \[\frac{{{\pi }^{4}}}{96}\]
B) \[\frac{{{\pi }^{4}}}{45}\]
C) \[\frac{89}{90}{{\pi }^{4}}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{4}^{4}}}+.....\infty \]\[=\frac{{{\pi }^{4}}}{90}\] \[\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+...+\infty +\frac{1}{{{2}^{4}}}\left( \frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{4}^{4}}}+...\infty \right)\] \[=\frac{{{\pi }^{4}}}{90}\] \[\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+\frac{1}{{{7}^{4}}}+......+\infty \]\[+\frac{1}{16}\times \frac{{{\pi }^{4}}}{90}=\frac{{{\pi }^{4}}}{90}\] \ \[\frac{1}{{{1}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{5}^{4}}}+\frac{1}{{{7}^{4}}}+.....+\infty \]\[=\frac{{{\pi }^{4}}}{90}-\frac{1}{16}\left( \frac{{{\pi }^{4}}}{90} \right)\] \[=\frac{15}{16}\left( \frac{{{\pi }^{4}}}{90} \right)=\frac{{{\pi }^{4}}}{96}\].
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