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Question Number 216926 by Engr_Jidda last updated on 24/Feb/25

$$Evaluate{5}^2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2}{\left(\underset{m=2}{\sum ^{\mathrm{\infty }}}\frac{2}{m^2+2m}\right)}^{n-1}$$

Answered by Wuji last updated on 25/Feb/25

$$5^2\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2}{\left(\underset{\mathrm{m}=2}{\sum ^{\mathrm{\infty }}}\frac{2}{{\mathrm{m}}^2+2\mathrm{m}}\right)}^{\mathrm{n}-1}$$$$\underset{\mathrm{m}=2}{\sum ^{\mathrm{\infty }}}\frac{2}{{\mathrm{m}}^2+2\mathrm{m}}=\underset{\mathrm{m}=2}{\sum ^{\mathrm{\infty }}}\frac{2}{\mathrm{m}(\mathrm{m}+2)}=\underset{\mathrm{m}=2}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{m}}-\frac{1}{\mathrm{m}+2})=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$$$5^5\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2}{\left(\frac{5}{6}\right)}^{\mathrm{n}-1}=25\bullet \frac{1}{2}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{5}{6}\right)}^{\mathrm{n}-1}=\frac{25}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{5}{6}\right)}^{\mathrm{n}}$$$$\mathrm{from}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{r}}^{\mathrm{n}}=\frac{1}{1-\mathrm{r}}\mathrm{for}\mid \mathrm{r}\mid <1$$$$\frac{25}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{5}{6}\right)}^{\mathrm{n}}=\left(\frac{25}{2}\right)\frac{1}{1-\frac{5}{6}}=\frac{25}{2}\left(6\right)=25\times 3=75$$$$$$

Commented by Engr_Jidda last updated on 25/Feb/25

$$thanks$$

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