Question

等差数列﹛an﹜的前n项和为Sn,且满足a3=10,S7=91.数列﹛bn﹜的前n项和Tn满足Tn=2bn-2恒成立 n

Answer 1

1、

a1+2d = 10;

7(a1+a1+6d)/2 = 91

解得：a1= 4,d=3

an = 3n+1.

Tn = 2bn-2;

T(n-1)=2b(n-1)-2;

Tn-T(n-1) = 2bn-2b(n-1)=bn;

bn = 2b(n-1),b1=2;

bn = 2^n.

2、

去除的项为b1 b4 b7.b(3n-2)为公比=8的等比数列

其和为Bn = 2(8^n-1)/7

数列b1,b2.b(3n)中,有b1,b4,.b(3n-2)

则A(2n) = 2(8^n-1) - 2(8^n-1)/7 = 12(8^n-1)/7 ①

数列b1,b2.b(3n-1)中,有b1,b4,.b(3n-2)

故A(2n-1) = 5*(8^n-1)/7 - 1 ②

所以A(2n) = 12(8^n-1)/7

A(2n-1) = 5(8^n-1)/7 -1

3、

an = 3n+1

bm = 2^m

3n+1 = 2^m

当n=1,m=2时,等式成立,说明这样的n和m是存在的.

3n = 2^m-1

n = (2^m-1)/3 (m∈N＋)