1 - 1-3 1 - 1-4 1 - 1-5 - 1 - 1-n equals

The correct option is C 2/n ( 1 13 ) ( 1 14 ) ( 1 15 )... ( 1 1n ) = ( 23 ) ( 34 ) (45 )... ( n 1n ) =2n ...

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Series

1 - 1/3 1 - ...

Question

$(1 - \frac{1}{3}) (1 - \frac{1}{4}) (1 - \frac{1}{5}) . . . (1 - \frac{1}{n})$ equals

\frac{1}{n}

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\frac{2}{n}

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\frac{3}{n}

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\frac{4}{n}

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$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

+ i v e

n

D =

∣ ∣ ∣ ∣ ∣ \begin{matrix} (n - 1)! & (n + 2)! & (n + 3)! / n (n + 1)! (n + 1)! & (n + 3)! & (n + 5)! / (n + 2)! (n + 3)! (n + 3)! & (n + 5)! & (n + 7)! / (n + 4)! (n + 5)! \end{matrix} ∣ ∣ ∣ ∣ ∣

\frac{D}{(n - 1)! (n + 1)! (n + 3)!}

lim n \to \infty ⎡ ⎢ ⎣ ⎧ ⎨ ⎩ 1 + {(\frac{1}{n})}^{4} ⎫ ⎬ ⎭ {⎧ ⎨ ⎩ 1 + {(\frac{2}{n})}^{4} ⎫ ⎬ ⎭}^{1 / 2} {⎧ ⎨ ⎩ 1 + {(\frac{3}{n})}^{4} ⎫ ⎬ ⎭}^{1 / 3} {⎧ ⎨ ⎩ 1 + {(\frac{4}{n})}^{4} ⎫ ⎬ ⎭}^{1 / 4} . . . . 2^{1 / n} ⎤ ⎥ ⎦

n \sum r = 1 T_{r} = (3^{n} - 1),

n \sum r = 1 \frac{1}{T_{r}}

\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + . . . . . + \frac{1}{n (n + 1) (n + 2)} = \frac{n (n + 3)}{4 (n + 1) (n + 2)}

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