已知等差数列{an}的前n项和为Sn,且a2=3,S15=225,设bn=2的an次方-2n,求数列{bn}的前n项和Tn
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已知等差数列{an}的前n项和为Sn,且a2=3,S15=225,设bn=2的an次方-2n,求数列{bn}的前n项和Tn
已知等差数列{an}的前n项和为Sn,且a2=3,S15=225,设bn=2的an次方-2n,求数列{bn}的前n项和Tn
已知等差数列{an}的前n项和为Sn,且a2=3,S15=225,设bn=2的an次方-2n,求数列{bn}的前n项和Tn
设{an}的公差为 d ,则
a1+d=3 ,----------①
15a1+105d=225 ,----------②
解得 a1=1,d=2 ,
因此 an=2n-1 .
由 bn=2^(2n-1)-2n
得 Tn=[2^1+2^3+.+2^(2n-1)]-2(1+2+...+n)
=2(1-4^n)/(1-4)-n(n+1)
=2/3*(4^n-1)-n(n+1) .
an=a1+(n-1)d
a2= a1+d=3 (1)
S15= a1+7d=15 (2)
(2)-(1)
6d=12
d=2
a1=1
an=1+(n-1)2= 2n-1
bn=2^(an)-2n
= 2^(2n-1) -2n
Tn =b1+b2+...+bn
= (2/3)(2^(2n)-1) - n(n+1)
∵等差数列{an}的前n项和为Sn,且a2=3,S15=225
∴15a8=225,得a8=15
则公差d=(15-3)/(8-2)=2
所以an=1+2(n-1)=2n-1
则bn=2^(2n-1) - 2n=1/2 4^n-2n
故Tn=1/2[4(1-4^n)/(1-4)]-2n(n+1)/2
=2/3(4^n-1)-n(n+1)
设an=a1+(n-1)d, sn=na1+n(n-1)d/2.则
a1+d=3 15a1+105d=225
∴ a1=1 d=2
∴bn=2^[1+2(n-1)]-2n=2^(2n-1)-2n
而2+2^3+2^5+....+2^(2n-1)=[2^(2n+1)-2]/3
-2-4-6-...-2n=-n(n+1)
∴Tn=[2^(2n+1)-2]/3-n(n+1)