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Let-n-be-positive-integer-satisfies-a-n-1-1-n-1-n-1-1-n-1-n-1-Find-the-value-of-a-1-a-2-a-3-a-99-




Question Number 209881 by naka3546 last updated on 24/Jul/24
Let n be positive integer satisfies    a_n  = 1 + (√(1/n)) − (√(1/(n+1))) − (√((1/n) − (1/(n+1))))    Find  the  value  of        a_1 a_2 a_3  …a_(99)
$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{satisfies} \\ $$$$ \\ $$$${a}_{{n}} \:=\:\mathrm{1}\:+\:\sqrt{\frac{\mathrm{1}}{{n}}}\:−\:\sqrt{\frac{\mathrm{1}}{{n}+\mathrm{1}}}\:−\:\sqrt{\frac{\mathrm{1}}{{n}}\:−\:\frac{\mathrm{1}}{{n}+\mathrm{1}}} \\ $$$$ \\ $$$$\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\:\mathrm{of}\: \\ $$$$ \\ $$$$\:\:\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} \:\ldots{a}_{\mathrm{99}} \\ $$
Commented by Frix last updated on 24/Jul/24
I found this:  Π_(k=1) ^(n(n+2)) (a_k ) =(2/((n+2)(n+1)))  99=n(n+2) ⇒ n=9 ⇒ answer is (1/(55))
$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}\left({n}+\mathrm{2}\right)} {\prod}}\left({a}_{{k}} \right)\:=\frac{\mathrm{2}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)} \\ $$$$\mathrm{99}={n}\left({n}+\mathrm{2}\right)\:\Rightarrow\:{n}=\mathrm{9}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{55}} \\ $$
Commented by naka3546 last updated on 24/Jul/24
Thanks
$$\mathrm{Thanks} \\ $$

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