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Question Number 55094 by naka3546 last updated on 17/Feb/19

$$\frac{\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+...}{1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+...}=x$$$$3x^2+2x-1=?$$

Commented by maxmathsup by imad last updated on 17/Feb/19

$$wehave\frac{1}{4}+\frac{1}{16}+\frac{1}{16}+...=1+\left(\frac{1}{4}\right)+{\left(\frac{1}{4}\right)}^2+....-1$$$$=\frac{1}{1-\frac{1}{4}}-1=\frac{4}{3}-1=\frac{1}{3}$$$$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+...=1+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{5}\right)}^2+{\left(\frac{1}{7}\right)}^2+...=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}$$$$but\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}$$$$\Rightarrow (1-\frac{1}{4})\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}\Rightarrow \frac{3}{4}\frac{{\pi }^2}{6}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=\frac{{\pi }^2}{8}\Rightarrow x=\frac{\frac{1}{3}}{\frac{{\pi }^2}{8}}=\frac{8{\pi }^2}{3}\Rightarrow$$$$3x^2+2x-1=3{\left(\frac{8{\pi }^2}{3}\right)}^2+2\left(\frac{8{\pi }^2}{3}\right)-1=\frac{64{\pi }^4}{3}+\frac{16{\pi }^2}{3}-1.$$

Commented by maxmathsup by imad last updated on 17/Feb/19

$$errorfromline6x=\frac{8}{3{\pi }^2}\Rightarrow$$$$3x^2+2x-1=3{\left(\frac{8}{3{\pi }^2}\right)}^2+2\frac{8}{3{\pi }^2}-1=\frac{64}{3{\pi }^4}+\frac{16}{3{\pi }^2}-1.$$

Answered by Joel578 last updated on 17/Feb/19

Answered by tm888 last updated on 18/Feb/19

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