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Factorial pattern

Source: Twenty-six Years of Problem Posing by John Mason

1 \times 1! = 2! - 1

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

1 \times 1! + 2 \times 2! + 3 \times 3! = 4! - 1

Does this pattern continue?

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Yes. It can be shown by induction:

First it's easy to check that

1 \times 1! = 2! - 1

. Next assume that

1 \times 1! + 2 \times 2! + . . . + k \times k! = (k + 1)! - 1

. Now we try to show the pattern holds for

k + 1

.

1 \times 1! + 2 \times 2! + . . . + k \times k! + (k + 1) \times (k + 1)!

= (k + 1)! - 1 + (k + 1) \times (k + 1)!

= (1 + k + 1) (k + 1)! - 1

= (k + 2)! - 1

Hence, by induction, the pattern holds for all

k \geq 1

.

Tags: numbers, factorials

If you enjoyed this puzzle, check out Sunday Afternoon Maths LVIII,
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