已知数列{an}前n项和为sn=2^(n+2)-4 设Bn=an*log2(an) 求数列b的前n项和
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已知数列{an}前n项和为sn=2^(n+2)-4 设Bn=an*log2(an) 求数列b的前n项和
已知数列{an}前n项和为sn=2^(n+2)-4 设Bn=an*log2(an) 求数列b的前n项和
已知数列{an}前n项和为sn=2^(n+2)-4 设Bn=an*log2(an) 求数列b的前n项和
a1=s1=2^(1+2)-4=4
sn=2^(n+2)-4
s(n-1)=2^(n+1)-4
an=sn-s(n-1)
an=2^(n+2)-4-[2^(n+1)-4]
=2^(n+2)-2^(n+1)
=2*2^(n+1)-2^(n+1)
=2^(n+1)
an=2^(n+1)
当n=1时,也符合
Bn=an*log2(an)
=2^(n+1)log2[2^(n+1)]
=(n+1)*2^(n+1)
b1=2*2^2
b2=3*2^3
.
bn=(n+1)*2^(n+1)
Sbn=b1+b2+b3+.+bn
=2*2^2+3*2^3+.+(n+1)*2^(n+1)
2Sbn=2*2^3+3*2^4+.+(n+1)^(n+2)
Sbn-2Sbn=2*2^2+2^3+2^4+.+2^(n+1)-(n+1)*2^(n+2)
-Sbn=8+8*[1-2^(n-1)]/(1-2)-(n+1)*2^(n+2)
-Sbn=8+8*[2^(n-1)-1]-(n+1)*2^(n+2)
-Sbn=8+8*2^(n-1)-8-(n+1)*2^(n+2)
-Sbn=8*2^(n-1)-(n+1)*2^(n+2)
-Sbn=2^(n+2)-(n+1)*2^(n+2)
Sbn=(n+1)*2^(n+2)-2^(n+2)
Sbn=(n+1-1)*2^(n+2)
Sbn=n*2^(n+2)