数学高1的
数列{an}为等比数列,Tn=n a1+(n-1)a2+....+2 an-1+an已知T1=1,T2=4 1,求数列{an}的首相和公比 2求数列{Tn}的通项公式
(1)设:{an} = {a1*q^n|n=正整数} T1 = a1 = 1 T2 = 2a1+a2 = 2a1+a1q = 4 ;===》q = 2 (2) Tn - T(n-1) = a1+a2+...+an = 2^n - 1 T(n-1) - T(n-2) = a1+a2+...+a(n-1) = 2^(n-1) - 1 ..... T2 - T1 = a1 + a2 = 2^2 - 1 上面式子加和 得:Tn - T1 = 2^(n+1) - n - 3 所以: Tn = 2^(n+1) - n - 2
答:{an}为等比数列,Tn=na1+(n-1)a2+……+2a(n-1)+an (1)已知T1=1, 则有 T1=a1=1, T2=4,则有 T2=2*a1+(2...详情>>
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