$1.1! + 2.2! + 3.3! + . . . . . + 10.10! =$

11!

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11! + 1

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11! - 1

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11.10!

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Solution

The correct option is C $11! - 1$
$T_{n} = n \cdot n!$
$= (n + 1)! - n!$
$S_{10} = 1.1! + 2.2! + 3.3! + \dots + 10.10!$
$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + (11! - 10!)$
$= 11! - 1!$
$= 11! - 1$

Hence, option C.

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Similar questions

n! = 1 \times 2 \times 3 \dots \dots n

p = 1.1! + (2.2!) + (3.3!) + \dots \dots + (10.10!)

lim n \to \infty \frac{1.1! + 2.2! + 3.3! + \dots + n . n!}{(n + 1)!} =

lim n \to \infty \frac{1.1! + 2.2! + 3.3! + . . . + n . n!}{(n + 1)!} =

1 + 1.1! + 2.2! + 3.3! + . . . + n . n!

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