Evaluate-5-2-n-1-1-2-m-2-2-m-2-2m-n-1-

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

E v a l u a t e 5^{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{2} {(\underset{m = 2}{\sum^{\infty}} \frac{2}{m^{2} + 2 m})}^{n - 1}

Answered by Wuji last updated on 25/Feb/25

5^{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{2} {(\underset{m = 2}{\sum^{\infty}} \frac{2}{m^{2} + 2 m})}^{n - 1}

\underset{m = 2}{\sum^{\infty}} \frac{2}{m^{2} + 2 m} = \underset{m = 2}{\sum^{\infty}} \frac{2}{m (m + 2)} = \underset{m = 2}{\sum^{\infty}} (\frac{1}{m} - \frac{1}{m + 2}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}

5^{5} \underset{n = 1}{\sum^{\infty}} \frac{1}{2} {(\frac{5}{6})}^{n - 1} = 25 ∙ \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(\frac{5}{6})}^{n - 1} = \frac{25}{2} \underset{n = 0}{\sum^{\infty}} {(\frac{5}{6})}^{n}

from \underset{n = 0}{\sum^{\infty}} r^{n} = \frac{1}{1 - r} for ∣ r ∣< 1

\frac{25}{2} \underset{n = 0}{\sum^{\infty}} {(\frac{5}{6})}^{n} = (\frac{25}{2}) \frac{1}{1 - \frac{5}{6}} = \frac{25}{2} (6) = 25 \times 3 = 75

Commented by Engr_Jidda last updated on 25/Feb/25

t h a n k s