Question

己知等差数列{An}的首项为１，前10的和为145,求A2+A4+...A2n

Answer 1

等差数列前10项的和S10=10*a1+10*(10-1)*d/2
145=10*1+10*9*d/2,∴d=3
an=3n-2

a2+a4+a8+……+a(2^n)
=3*2-2+3*4-2+3*8-2+……+3*2^n-2
=3*（2+4+8+……+2^n）-2n 2，4，8，……，2^n等比
=3*[2(1-2^n)/(1-2) ]-2n
=6*2^n-2n-6

Answer 2

a(1)=1
sn=n*a(1)+1/2*n*(n-1)d
则s10=10a(1)+1/2*10*9*d=145
则10+45d=145
则d=3

故a(2)=a(1)+d=1+3=4

因为：A(2)+A(4)+...A(2n)
实质上是首项a(2),公差为2d的共有n项的等差数列前n项和。
所以：A(2)+A(4)+...A(2n)=n*a(2)+n/2*(n-1)*2d.
=4n+n/2*(n-1)*6
=4n+3n(n-1)
=3n^2+n