If $1 \cdot (0!) + 3 \cdot (1!) + 7. (2!) + 13 \cdot (3!) + 21 \cdot (4!) + . . .$ upto n terms $= (4000) 4000!$ , then the value of n is

4000

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4001

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3999

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None of these

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Solution

The correct option is C 3999
Given,

$1. (0!) + 3. (1!) + 7. (2!) + . . . . .$ upto $n$ terms $= (4000) 4000!$

we have a formula,

$\sum [k! [(k + 1) (k + 2) - 2 (k + 1) + 1] = (k + 1) [(k + 1)!]$

Since the given result is $(4000) (4000!)$
Hence the k value is

$k = 3999$

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1. (0!) + 3. (1!) + 7. (2!) + 13. (3!) + 21. (4!) + . . .

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= (4000) 4000!

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(4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + \dots u p t o n t e r m s

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