Question

设{an}为等差数列，Sn为数列{an}的前n项和，已知S7=7,S15=75设{an}为等差数列，Sn为数列{an}的前n项和

设{an}为等差数列，Sn为数列{an}的前n项和，已知S7=7,S15=75
设{an}为等差数列，Sn为数列{an}的7,S15=75（1）求数列{an}的通项公式
2)设bn=2^an求{n*bn}的前n项和Tn

Answer 1

a(n) = a+(n-1)d.
s(n) = na + n(n-1)d/2.

7 = s(7) = 7a + 21d, 1 = a + 3d.
75 = s(15) = 15a + 15*7d = 15(a+7d), 5 = a + 7d.
4 = 5-1 = (a+7d) - (a+3d) = 4d. d = 1.
a = 1 - 3d = -2.

a(n) = -2 + (n-1) = n-3.

b(n) = 2^[a(n)] = 2^(n-3).

t(n) = 1*b(1)+2*b(2)+3*b(3) + ... + (n-1)b(n-1)+nb(n)
= 2^(1-3) + 2*2^(2-3) + 3*2^(3-3) + ... + (n-1)2^(n-4) + n2^(n-3),

2t(n) = 2^(2-3) + 2*2^(3-3) + ... + (n-1)2^(n-3) + n2^(n-2).

t(n) = 2t(n) - t(n) = -2^(1-3) - 2^(2-3) - 2^(3-3) - ... - 2^(n-3) + n2^(n-2)
= n2^(n-2) -(1/4)[1 + 1/2 + ... + 1/2^(n-1)]
= n2^(n-2) - (1/4)[1-1/2^n]/(1-1/2)
= n2^(n-2) - [1-1/2^n]/2
= n2^(n-2) - 1/2 + 1/2^(n+1)