已知数列{an}中,a1=3,且a的n+1项=2a的n项-1(1)求{an}的通项公式(2)求数列{nan}前n项和(3
1个回答
(1)
a(n+1) = 2an -1
a(n+1)-1 = 2(an -1)
[a(n+1)-1]/(an -1)=2
(an -1)/(a1 -1)=2^(n-1)
an = 1+ 2^n
(2)
cn =nan
= n + 2.(n.2^(n-1)
Tn = c1+c2+...+cn
= n(n+1)/2 + 2[∑(i:1->n) i.2^(i-1)]
consider
1+x+x^2+...+x^n = (x^(n+1) -1)/(x-1)
1+2x+...+nx^(n-1) = [(x^(n+1) -1)/(x-1)]'
= (nx^(n+1) -(n+1)x^n +1)/(x-1)^2
put x=2
∑(i:1->n) i.2^(i-1) = n.2^(n+1) -(n+1).2^n +1
= 1+ (n-1).2^n
Tn = c1+c2+...+cn
= n(n+1)/2 + 2[∑(i:1->n) i.2^(i-1)]
=n(n+1)/2 + 2[1+ (n-1).2^n]
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