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Question Number 201408 by Ari last updated on 05/Dec/23

Answered by mr W last updated on 05/Dec/23

$$=\frac{1\times 3}{2^2}\times \frac{2\times 4}{3^2}\times \frac{3\times 5}{4^2}\times ...\times \frac{98\times 100}{{99}^2}$$$$=\frac{1\times \cancel{3}}{2^{\cancel{2}}}\times \frac{\cancel{2}\times \cancel{4}}{\cancel{3^2}}\times \frac{\cancel{3}\times \cancel{5}}{\cancel{4^2}}\times ...\times \frac{\cancel{98}\times 100}{{99}^{\cancel{2}}}$$$$=\frac{1}{2}\times \frac{100}{99}$$$$=\frac{50}{99}$$

Answered by Mathspace last updated on 05/Dec/23

$$\prod _{k=2}^n(1-\frac{1}{k^2})=A_n$$$$\Rightarrow A_n=\prod _{l=2}^n\frac{(k-1)(k+1)}{k^2}$$$$=\prod _{k=2}^n\frac{k-1}{k}\times \frac{k+1}{k}$$$$=\frac{1}{2}\times \frac{3}{2}\times \frac{2}{3}\times \frac{4}{3}\times ...\times \frac{n-2}{n-1}\times \frac{n}{n-1}\times$$$$\frac{n-1}{n}\times \frac{n+1}{n}=\frac{1}{2}=\frac{n+1}{2n}$$$$A_{99}=\frac{99+1}{2\times 99}=\frac{100}{2.99}=\frac{50}{99}$$

Answered by tri26112004 last updated on 06/Dec/23

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