a2=a1+d=4
S10=10a1+45d=10a1+10d+35d=10(a1+d)+35d=40+35d=145
35d=105
d=3
a1=a2-d=4-d=4-3=1
数列{an}是以1为首项,3为公差的等差数列.
an=1+3(n-1)=3n-2
①
a2+a4+a8+...+a(2^n)
=a1+2d-d+a1+4d-d+a1+8d-d+...+a1+(2^n)d-d
=na1+d(2+4+8+...+2^n)-nd
=na1+2(2^n-1)d/(2-1)-nd
=na1+2d(2^n-1)-nd
=1×n+2×3×(2^n-1)-3×n
=n+6×2^n-6-3n
=3×2^(n+1)-2n-6
②
2^n×an=2^n×(3n-2)=3n×2^n-2^(n+1)
2a1+4a2+8a3+...+2^n×an=3×(1×2^1+2×2^2+...+n×2^n)-[2^2+2^3+...+2^(n+1)]
令Cn=1×2^1+2×2^2+...+n×2^n
则2Cn=1×2^2+2×2^3+...+(n-1)×2^n+n×2^(n+1)
Cn-2Cn=-Cn=2^1+2^2+2^3+...+2^n-n×2^(n+1)
=2(2^n-1)/(2-1)
=2^(n+1)-2-n×2^(n+1)
=(1-n)×2^(n+1)-2
Cn=(n-1)×2^(n+1)+2
2a1+4a2+8a3+...+2^n×an=3×(1×2^1+2×2^2+...+n×2^n)-[2^2+2^3+...+2^(n+1)]
=3[(n-1)×2^(n+1)+2]-4(2^n-1)/(2-1)
=(3n-3)×2^(n+1)+6-2×2^(n+1)+4
=(3n-5)×2^(n+1)+10