- n -1-1- n -1 n -2 n -3 - - n - k is equal toA- 1- k -1 k -1B- 1- k k - kC- 1- k -1 1- kD- 1-k

The correct option is C T_n=1/(n+1)(n+2)(n+3)....(n+k) V_n 1=1/(n+1)(n+2)(n+3)....(n+k 2)(n+k 1) V_n V_n 1=1/(n+2)(n+3)....(n+k 1)[1/n+k 1/n+1] =1 k/(n+1)(n+2 ...

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Byju's Answer

Standard XII

Mathematics

Constant Function

∑ n =1∞1/ n +...

Question

$\infty \sum n = 1 \frac{1}{(n + 1) (n + 2) (n + 3) . . . . (n + k)}$ is equal to

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Solution

The correct option is C

$T_{n} = \frac{1}{(n + 1) (n + 2) (n + 3) . . . . (n + k)}$

$V_{n - 1} = \frac{1}{(n + 1) (n + 2) (n + 3) . . . . (n + k - 2) (n + k - 1)}$

$V_{n} - V_{n - 1} = \frac{1}{(n + 2) (n + 3) . . . . (n + k - 1)} [\frac{1}{n + k} - \frac{1}{n + 1}]$

$= \frac{1 - k}{(n + 1) (n + 2) (n + 3) . . . . (n + k - 1) (n + k)}$

$V_{n} - V_{n - 1} = (1 - k) T_{n}$

$∴ (1 - k) T_{n} = V_{n} - V_{n - 1}$

$(1 - k) T_{1} = V_{1} - V_{0}$

$(1 - k) T_{2} = V_{2} - V_{1}$

$(1 - k) T_{3} = V_{3} - V_{2}$

$. . . . . . . . . . . . . . . . . . . . . . . . . . .$

$(1 - k) T_{n} = V_{n} - V_{n - 1}$

$- - - - - - - - - - - - - - - - - - -$

Adding $1 - k) S_{n} = V_{n} - V_{0}$

or $(1 - k) S_{n} = \frac{1}{(n + 1) (n + 2) . . . . . . (n + k)} - \frac{1}{1.2.3.... k} (1 - k)_{n \to \infty}^{L t} S_{n} = 0 - \frac{1}{k!}$

$_{n \to \infty}^{L t} S_{n} = \frac{1}{(k - 1) k!}$

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