When simplified- the product 1- 1 3 1- 1 4 1- 1 5 - 1- 1n becomes

The correct option is B 2 n ( 1 1 3 #160; ) ( 1 1 4 #160; ) ( 1 1 5 #160; ) ...( 1 1n ) = 2 3 . 3 4 . 4 5 .... n 2 n 1 . n 1 ...

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Bayes' Theorem

When simplifi...

Question

When simplified, the product $(1 - \frac{1}{3}) (1 - \frac{1}{4}) (1 - \frac{1}{5}) . . . (1 - \frac{1}{n})$ becomes

\frac{1}{n}

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\frac{2}{n}

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\frac{2 (n - 1)}{n}

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\frac{2}{n (n + 1)}

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Solution

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Similar questions

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1 \times n + 2 (n - 1) + 3 (n - 2) + \dots \dots + (n - 1) \times 2 + n \times 1

$n^{n} - n (n - 1)^{n} + \frac{n (n - 1)}{2!} (n - 2)^{n} - . . . \infty = \frac{3}{a} \times n!$ .

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6

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

\frac{3}{1 \cdot 2} (\frac{1}{2}) + \frac{4}{2 \cdot 3} {(\frac{1}{2})}^{2} + \frac{5}{3 \cdot 4} {(\frac{1}{2})}^{3} + \dots

n

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