If n is a positive integer- then the value of n - 4-n - 7- is-

The correct option is D 1(n + 5)(n + 6)(n + 7) The value of (n+4)! (n+7)! = (n+4)! (n+7)(n+6)(n+5)(n+4)! = 1 (n+7)(n+6)(n+5) ...

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Solving Linear Equations Using Transpose of Terms

If n is a p...

Question

If $n$ is a positive integer, then the value of $\frac{(n + 4)!}{(n + 7)!}$ is:

\frac{1}{(n + 1) (n + 2) (n + 3)}

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\frac{1}{n + 5}

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\frac{1}{(n + 5) (n + 6)}

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\frac{1}{(n + 6) (n + 7)}

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\frac{1}{(n + 5) (n + 6) (n + 7)}

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3

1 - \frac{n - 3}{2} + \frac{(n - 4) (n - 5)}{⌊ 3} - \frac{(n - 5) (n - 6) (n - 7)}{⌊ 4} + . . . . .

+ (- 1)^{r - 1} \frac{(n - r - 1) (n - r - 2) . . . . (n - 2 r + 1)}{⌊ r} + . . . . . .,

\frac{3}{n}

- \frac{1}{n}

n

D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (n - 1)! & (n + 1)! & (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 [\frac{n}{(n + 1) (n + 2)} + \frac{n}{(n + 2) (n + 4)} + \dots + \frac{1}{6 n}]

S_{n} = \frac{n}{(n + 1) (n + 2)} + \frac{n}{(n + 2) (n + 4)} + \frac{n}{(n + 3) (n + 6)} + . . . . . . + \frac{1}{6 n}

lim n \to \infty S_{n}

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