Question

已知{an}是等差数列,a3=7,a5+a7=26.Sn为{an}的前n项和①求通项an及Sn②设

已知{an}是等差数列,a3=7,a5+a7=26.Sn为{an}的前n项和①求通项an及Sn②设{bn-an}是首项为1,公比为2的等比数列,求数列{bn}的通项公式及其前n项和Tn

Answer 1

设{an}公差为d
1.
a5+a7=2a6=26
a6=13
a6-a3=3d=13-7=6
d=2
a1=a3-2d=7-2×2=3
an=a1+(n-1)d=3+2(n-1)=2n+1
Sn=(a1+an)n/2=(3+2n+1)n/2=n²+2n
数列{an}的通项公式为an=2n+1，前n项和Sn=n²+2n
2.
bn-an=1×2^(n-1)=2^(n-1)
bn=an+2^(n-1)=2^(n-1)+2n+1
Tn=b1+b2+...+bn
=[1+2+...+2^(n-1)]+Sn
=1×(2ⁿ-1)/(2-1)+n²+2n
=2ⁿ+n²+2n -1
数列{bn}的通项公式为bn=2^(n-1)+2n+1，前n项和Tn=2ⁿ+n²+2n -1