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}前n项的和Tn.

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