3.5.1 1 · 1! + 2 · 2! + · · · + n · n!

The sum of 1⋅1!+2⋅2!+⋯+n⋅n!1 \cdot 1! + 2 \cdot 2! + \dots + n \cdot n! is a fairly simple problem if you know the formula (its derivation is left to the reader).

1⋅1!+2⋅2!+⋯+n⋅n!=(n+1)!−11 \cdot 1! + 2 \cdot 2! + \dots + n \cdot n! = (n + 1)! - 1

The simplest case would be to compute sums like:

1⋅1!+2⋅2!+3⋅3!+4⋅4!=(4+1)!−1=120−1=1191 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + 4 \cdot 4! = (4 + 1)! - 1 = 120 - 1 = 119

There are slight variations which could be asked (the easiest of which would be leaving out some terms).

1⋅1!+3⋅3!+5⋅5!=(5+1)!−1−2⋅2!−4⋅4!=720−1−4−96=6191 \cdot 1! + 3 \cdot 3! + 5 \cdot 5! = (5 + 1)! - 1 - 2 \cdot 2! - 4 \cdot 4! = 720 - 1 - 4 - 96 = 619

The following are some practice problems:

Problem Set

1cdot1!+2cdot2!+3cdot3!+4cdot4!+5cdot5!=1 \\cdot 1! + 2 \\cdot 2! + 3 \\cdot 3! + 4 \\cdot 4! + 5 \\cdot 5! =
1cdot1!+2cdot2!+dots+6cdot6!=1 \\cdot 1! + 2 \\cdot 2! + \\dots + 6 \\cdot 6! =
1cdot1!+2cdot2!+dots+7cdot7!=1 \\cdot 1! + 2 \\cdot 2! + \\dots + 7 \\cdot 7! =
1cdot1!−2cdot2!−3cdot3!−4cdot4!=1 \\cdot 1! - 2 \\cdot 2! - 3 \\cdot 3! - 4 \\cdot 4! =
2cdot1!+3cdot2!+4cdot3!+5cdot4!=2 \\cdot 1! + 3 \\cdot 2! + 4 \\cdot 3! + 5 \\cdot 4! =