Question

高一数列题：已知数列an的前n项和为Sn,且满足Sn=2a-2

已知数列an的前n项和为Sn,且满足Sn=2an-2(n∈N+)在数列｛b n｝中，b1=1,点P(bn,bn+1)在直线x-y+2=0上，求： (1)求数列｛an｝,｛bn｝的通项公式 (2)记Tn=a1b1+a2b2+a3b3+…+anbn,求Tn

Answer 1

a(n+1)=s(n+1)-s(n)=2a(n+1)-2a(n),a(n+1)=2a(n)

a1=2a1-2,a1=2;
an=2^n;

b(n)-b(n+1)+2=0,b(n+1)=b(n)+2
b1=1
bn=2(n-1)+1=2n-1

anbn=(2n-1)*2^n
Tn-2Tn
=a1b1+a2b2+a3b3+…+anbn-2(a1b1+a2b2+a3b3+…+anbn)
=a1b1+(a2b2-2a1b1)+(a3b3-2a2b2)+...+(anbn-2a(n-1)b(n-1))-2anbn
=a1b1+()+()+...+((2n-1)*2^n-2(2(n-1)-1)*2^(n-1))-2anbn
=a1b1+()+()+...+((2n-1)*2^n-(2n-3)*2^n)-2anbn
=a1b1+()+()+...+(2*2^n)-2anbn
=a1b1+2^3+2^4+...+2^(n+1)-2anbn
=a1b1+2^(n+2)-2^3-2anbn
=2+2^(n+2)-8-2(2n-1)*2^n

Tn=-2-2^(n+2)+8+2(2n-1)*2^n

Answer 2