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n-0-2-n-2n-3-2n-1-n-1-n-Evaluate-




Question Number 195904 by York12 last updated on 13/Aug/23
Σ_(n=0) ^∞ [((2^n (2n)!)/(3^(2n+1) (n+1)!n!))]=λ  Evaluate (λ)
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{2}^{{n}} \left(\mathrm{2}{n}\right)!}{\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!{n}!}\right]=\lambda \\ $$$${Evaluate}\:\left(\lambda\right) \\ $$
Answered by qaz last updated on 13/Aug/23
(1/3)Σ_(n=0) ^∞  (((2n)),(n) )((2/9))^n (1/(n+1))=(1/3)Σ_(n=0) ^∞  (((−1/2)),((      n)) )(−4)^n ((2/9))^n ∫_0 ^1 x^n dx  =(1/3)∫_0 ^1 (1−(8/9)x)^(−1/2) dx=(1/2)
$$\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\left(\frac{\mathrm{2}}{\mathrm{9}}\right)^{{n}} \frac{\mathrm{1}}{{n}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{−\mathrm{1}/\mathrm{2}}\\{\:\:\:\:\:\:{n}}\end{pmatrix}\left(−\mathrm{4}\right)^{{n}} \left(\frac{\mathrm{2}}{\mathrm{9}}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{8}}{\mathrm{9}}{x}\right)^{−\mathrm{1}/\mathrm{2}} {dx}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by York12 last updated on 13/Aug/23
thanks
$${thanks} \\ $$

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