Maths prove

Bybinormial theorem:
(x + 1)^n - 1
= [x^n + nC1x^(n-1) + nC2x^(n-2) +...... + nCn-2x^2 + nCn-1­x + 1] -1
= x^n + nC1x^(n-1) + nC2x^(n-2) +...... + nCn-2x^2 + nCn-1­x
= x[x^(n-1) + nC1x^(n-2) + nC2x^(n-3)+ ...... + nCn-2x + nCn-1]

Hence, (x + 1)^n - 1 is divisible by x
and the quotient is x^(n-1) + nC1x^(n-2) + nC2x^(n-3)+ ...... + nCn-2x + nCn-1
wherex and n are positive integers.

2011-07-25 17:09:51 補充：
Alternative answer : (using mathematical induction)

Let P(n): (x + 1)^n - 1 is divisible by x, where x and n are positive integers.

When n = 1:
(x + 1)^1 - 1 = x
Hence, P(1) is true.

2011-07-25 17:10:40 補充：
Assume that P(k) is true, i.e. (x + 1)^k - 1 = xQ
where Q is the quotient when (x + 1)^k - 1 is divisible by x.
To prove that P(k + 1) is true, i.e. (x + 1)^(k + 1) - 1 is divisible by x.

2011-07-25 17:10:55 補充：
Proof:
(x + 1)^(k + 1) - 1
= (x + 1)(x + 1)^k - 1
= (x + 1)[(x + 1)^k - 1 + 1] - 1
= (x + 1)[(x + 1)^k - 1] + (x + 1) - 1
= (x + 1)xQ + x + 1 - 1
= (x + 1)xQ + x
= x[(x + 1)Q + 1]
Hence, P(k + 1) is true.

By the principle of mathematical induction, P(n) is true
when x and n are positive integers.

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