Question 15

If $$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + $$...... upto $$ \infty = \frac{\pi^{2}}{6}$$, then value of $$\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + $$...... upto $$\infty$$ is

$$\dfrac{1}{1^{2}} + \dfrac{1}{2^{2}} + \dfrac{1}{3^{2}} + $$...... upto $$ \infty = \dfrac{\pi^{2}}{6}$$

the above equation can be written as :
[$$\dfrac{1}{1^{2}} + \dfrac{1}{3^{2}} + \dfrac{1}{5^{2}} + $$......$$upto\ \infty\ $$] + [$$\dfrac{1}{2^{2}} + \dfrac{1}{4^{2}} + $$......$$upto\ \infty$$] = $$\dfrac{\pi^{2}}{6}$$

Now lets assume $$\dfrac{1}{1^{2}} + \dfrac{1}{3^{2}} + \dfrac{1}{5^{2}} + $$......$$upto\ \infty\ $$ = X.

And for the second half of the equation lets take $$\ \dfrac{\ 1}{2^2}$$ as common :

X + $$\ \dfrac{\ 1}{2^2}$$[$$\dfrac{1}{1^{2}} + \dfrac{1}{2^{2}} + \dfrac{1}{3^{2}} + $$......$$upto\ \infty\ $$ ] = $$\dfrac{\pi^{2}}{6}$$

X + $$\dfrac{\ 1}{4}$$[$$\dfrac{\pi^{2}}{6}$$] = $$\dfrac{\pi^{2}}{6}$$

Therefore, X = [$$\dfrac{\pi^{2}}{6}$$][$$\dfrac{\ 3}{4}$$] = $$\dfrac{\pi^{2}}{8}$$

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