標題:
MATHEMATICAL INDUCTION
發問:
Given that 1(3)+2(3^2)+3(3^3)+...+n(3^n)=[3+(2n-1)3^(n+1)]/4 is true for all positive integers n. Prove that 2(3)+3(3^2)+4(3^3)+...+21(3^20)=[41(3^21)-3)/4
最佳解答:
Method 1(Using G.P.Sum) 1(3)+2(3^2)+3(3^3)+......+n(3^n)=[3+(2n-1)3^(n+1)]/4 For n=20 1(3)+2(3^2)+3(3^3)+......+20(3^20)=[3+39(3^21)]/4---------(1) By using G.P Sum 3+3^2+3^3+......+3^20=[3(3^20-1)]/(3-1)=[3(3^20-1)]/2---------(2) 2(3)+3(3^2)+4(3^3)+...+21(3^20)=(1)+(2) =[3+39(3^21)]/4+[3(3^20)-3]/2 =[3+39(3^21)+6(3^20)-6]/4 =[39(3^21)+2(3^21)-3]/4 =[41(3^21)-3]/4 Method 2: (Direct calculation) 1(3)+2(3^2)+3(3^3)+......+n(3^n)=[3+(2n-1)3^(n+1)]/4 For n=21 1(3)+2^(3^2)+3(3^3)......+21(3^21)=[3+41(3^22)]/4 Divide the equation by 3 on both sides 1+2(3)+3(3^2)+......+21(3^20)=[1+41(3^21)]/4 2(2)+3(3^2)+......+21(3^20)=[1+41(3^21)]/4-1 ∴2(2)+3(3^2)+......+21(3^20)=[41(3^21)-3]/4 P.S. Difference in 2 methods are highlighted.
其他解答:
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