mathematical induction4.4

用戶36732

2009-10-07 5:04 am

(a)

Deduce a formula for evaluating Sn = 2-{1+(1/2)+(1/4)+(1/8)+...+1/[2^(n-1)]}, where n is a positive integer.

(b)

Prove, by mathematical induction, that the formula obtained in (a) is true for all positive integers n.

回答 (1)

用戶9112

2009-10-07 5:13 am

✔ 最佳答案

(a) Sn = 2 - {1+(1/2)+(1/4)+(1/8)+...+1/[2^(n-1)]}
= 2 - 1[1 - (1/2)^n] / (1 - 1/2)
= 2 - 2[1 - (1/2)^n]
= 1 / 2^(n - 1)
(b) Let P(n) be the proposition that Sn = 1 / 2^(n-1)
P(1) is true since S1 = 2 - 1 = 1 = 1 / 2^0
Assume P(k) is true, i.e. Sk = 1 / 2^(k - 1)
Sk+1 = 2 - {1+(1/2)+(1/4)+(1/8)+...+1/[2^(k-1)] + 1/2^k}
= Sk - 1/2^k
= 1 / 2^(k - 1) - 1 /2^k
= 1 / 2^k [2 - 1]
= 1 / 2^k
= 1/ 2^[(k+1) - 1]
=> P(k + 1) is also true
Hence by the principle of MI, P(n) is true for all positive integers n

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