Question

3.设数列{an}的前n项和Sn=2an-4(n∈N+),数列{bn}满足:bn+1=an+2bn,且b1=2.求{bn

Answer 1

1.

S(n)-S(n-1)=2(a(n)-a(n-1))=an

an=2a(n-1)

S1=2a1-4=a1====>a1=4,an=2的n+1次方

2.

bn+1=an+2bn=2bn+(2的n+1次方)

左右两边同除以2的n+1次方====〉

bn+1/2的n+1次方=[bn/(2的n次方)]+1

即bn/(2的n次方)为等差数列,首项为b1/2=1

公差为1

bn/(2的n次方)=1+n-1=n

bn=n*(2的n次方)

Sn-2Sn=2+2平方＋2(三方)＋...+2(n方)-n*2的n+1次方=(1-n)*2的n+1次方-2=-Sn

Sn=(n-1)2的n+1次方+2