1.1!+2.2!+3.3!+……+n.n!=

1.1!+2.2!+…+n⋅n!=∑k=1n k⋅(k!)=∑k=1n [k+1−1]k!=∑k=1n [(k+1)!−k!]=2!−1!+3!−2!+4!−3!+⋯ +(n+1)!−n!=(n+1)!−1

1.1!+2.2!+3.3!+……+n.n!=

1.1!+2.2!+…+n⋅n!=∑k=1n k⋅(k!)=∑k=1n [k+1−1]k!=∑k=1n [(k+1)!−k!]=2!−1!+3!−2!+4!−3!+⋯ +(n+1)!−n!=(n+1)!−1

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$1.1! + 2.2! + 3.3! + \dots \dots + n . n! =$

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$(n + 1)! - 1$

$(n + 1)!$

$(n + 1)! + 1$

$n (n + 1)!$

answer is A.

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$\begin{array}{l} 1 .1! + 2 .2! + \dots + n \cdot n! = \sum_{k = 1}^{n} k \cdot (k!) \\ = \sum_{k = 1}^{n} [k + 1 - 1] k! = \sum_{k = 1}^{n} [(k + 1)! - k!] \\ = 2! - 1! + 3! - 2! + 4! - 3! + \dots \\ + (n + 1)! - n! = (n + 1)! - 1 \end{array}$

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1.1!+2.2!+3.3!+……+n.n!=

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$1.1! + 2.2! + 3.3! + \dots \dots + n . n! =$