Lim n - k -1 n k3-6 k2-11 k -5- k -3 -A- 2-3B- 1- 5-3D- 7-3

The correct option is C 5/3k^3 + 6k^2 + 11k + 5 = (k+1) (k+2) (k+3) 1 ∴k^3 + 6k^2 + 11k + 5/(k+3)! = 1/k! 1/(k+3)! Therefore value is 5/3. ...

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Combination

lim n →∞∑ k =...

Question

$l i m n \to \infty \sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!} =$

\frac{2}{3}

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1

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\frac{5}{3}

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\frac{7}{3}

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13 \sum k = 1 \frac{1}{sin (\frac{π}{4} + \frac{(k - 1) π}{6}) sin (\frac{π}{4} + \frac{k π}{6})}

\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

k

6 k^{2}

2 \int 0 [x^{2}] d x =

x

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$P (X = x)$	$k$	$3 k$	$5 k$	$7 k$	$9 k$	$11 k$	$13 k$

^{'} k^{'}

¯ r = (3^i + 5^j + 7^k) + λ (^i - 2^j +^k)

¯ r = (-^i -^j -^k) + μ (7^i - 6^j +^k)

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