Question

(n + 1)! = n?
n² tn
Prove that
(n - 1)!​

Answer 1

Let’s construct the answer step by step.

1 + 1 + 1 + …. 1 (n times) = n

1 + 2 + 3 + … + n = (1/2) n^2 + (1/2)n

1^2 + 2^2 + 3^2 + … + n^2 = (1/3) n^3 + (1/2) n^2 + (1/6) n

1^3 + 2^3 + 3^3 + …. + n^3 = (1/4) n^4 + (1/2) n^3 + (1/4) n^2

Each formula is calculated by simple algorithm from previous formula.

To calculate the next formula from the current formula, do the following.

[1] Multiply the current formula by the next higher exponent which is also the reciprocal of the first coefficient in the current formula.

(1/3) n^3 → n^3

(1/2) n^2 → (3/2) n^2

(1/6) n → (1/2) n

[2] For each power of n term, increase the exponent on n by 1, and divide its coefficient by the new exponent.

n^3 → (1/4) n^4

(3/2) n^2 → (1/2) n^3

(1/2) n → (1/4) n^2

To find the coefficient of n, calculate the coefficient of n that will make the formula value equal to 1 when n = 1.

The formula for sum of cubes is

(1/4) n^4 + (1/2) n^3 + (1/4) n^2 + x n

When n = 1, this means that

(1/4) + (1/2) + (1/4) + x = 1

We calculate x = 0.

Thus the formula for the sum of n cubes, starting with 1^3 is

(1/4) n^4 + (1/2) n^3 + (1/4) n^2

You can confirm the validity of any of these formulas by [1] showing the correctness of (1 + exponent) numerical examples or [2] proving them by mathematical induction, or [3] by using difference equations.

Finally, to get the summation of (n^2+1) n = n^3 + n

we just add the formulas for the sum of n^3 and sum of n.

1 + 2 + 3 + … + n = (1/2) n^2 + (1/2)n

1^2 + 2^2 + 3^2 + … + n^2 = (1/3) n^3 + (1/2) n^2 + (1/6) n

1^3 + 2^3 + 3^3 + …. + n^3 = (1/4) n^4 + (1/2) n^3 + (1/4) n^2

to get

(1/4) n^4 + (1/2) n^3 + (3/4) n^2 + (1/2) n

(1/4) * 1^4 + (1/2) * 1^3 + (3/4) * 1^2 + (1/2) * 1 = 2 = (1^2 + 1)*1

(1/4) * 2^4 + (1/2) * 2^3 + (3/4) * 2^2 + (1/2) * 2 = 4 + 4 + 3 + 1 = 12

and (1^2 + 1) * 1 + (2^2 + 1) * 2 = 2 + 10 = 12