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討論串[中學] 數學歸納法
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前略 (sqrt(2)<3/2成立). 設S=2^(1/2^(2-1))*3(1/2^(3-1))*4(1/2^(4-1))*...*k^(1/2^(k-1)). <3/(k+2)^(1/2^(k-1))=T. 則S*(k+1)^(1/2^k)<T*(k+1)^(1/2^k). 這裡硬做會吃釘子.
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(c). check n = 0 is true. assume n = k, 1+x+x^2+ ... + x^k = [x^(k+1)-1]/(x-1) is true. n = k+1, 1+x+x^2+ ... + x^k + x^(k+1) = [x^(k+1)-1]/(x-1) + x^
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