proof-that-n-1-10-n-n-11-1-

Question Number 155803 by cortano last updated on 05/Oct/21

proof that

\underset{n = 1}{\sum^{10}} n \times n! = 11! - 1

Answered by som(math1967) last updated on 05/Oct/21

1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + 10 \times 10! + 1 - 1

= 2 + 2 \times 2! + 3 \times 3! + \dots + 10 \times 10! - 1

= 2! (1 + 2) + 3 \times 3! + 4 \times 4! + \dots + 10 \times 10! - 1

= 3! + 3 \times 3! + 4 \times 4! + \dots + 10 \times 10! - 1

= 3! (1 + 3) + 4 \times 4! + \dots + 10 \times 10! - 1

= 4! (1 + 4) + 5 \times 5! + \dots + 10 \times 10! - 1

= 5! (1 + 5) + \dots + 10 \times 10! - 1

= \dots

= 10! (1 + 10) - 1 = 11! - 1 [p r o v e d]

Answered by puissant last updated on 05/Oct/21

\underset{n = 1}{\sum^{10}} n \times n! = \underset{n = 1}{\sum^{10}} (n + 1 - 1) \times n!

= \underset{n = 1}{\sum^{10}} (n + 1) \times n! - n! = \underset{n = 1}{\sum^{10}} {(n + 1)! - n!}

= 2! - 1! + 3! - 2! + 4! - 3! + \dots . + 11! - 10!

= 11! - 1!

∴∵ \underset{n = 1}{\sum^{10}} n \times n! = 11! - 1 . .

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