Mathematical induction222

用戶36732

2009-09-27 7:22 pm

Let Sk = 1 - (1/2) + (1/3) - (1/4) + … + [-1^(k-1)] (1/k) ; Tn = [1/(n+1)] + [1/(n+2)] +…+ 1/(n + n), where k and n are positive integers.

(a) Express S2n in terms of n.
(b) Prove, by mathematical induction, that S2n = Tn for all positive integers n.

回答 (1)

So Much

2009-10-04 2:29 am

✔ 最佳答案

S_2n = 1-(1+1/2+1/3+...+1/2n)

PS. for the index of -1 , which present in the form of 2n-1 .

so the index must be odd . for any odd number k , -1^k=-1.

b)

When n=1,

S_2=1-1/2=1/2 = T_n , it is true for n =1 ,

Assume it is true for n=k ,where k is a positive integer ,

i.e. S_2k=T_k ,

Cosider n=k+1 ,

S_2(k+1)-T(k+1)

=1-(1/2+1/3+...+1/(2k+2))-[1/(k+2)+1/(k+3)+...+1/(2k+2)]

=S_2k-1/(2k+1)-1/(2k+2)-T_k+1/(2k+1)+1/(2k+2)

=0

so S_2(k+1)=T_(k+1) , it is true for n =k +1

By Mathematical induction , it is true for all positive integer n .

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