a1,a5,a17为等比数列
(a5)^2=a1*a17
(a1+4d)^2=a1(a1+16d)
16d^2-8a1d=0
a1=2d
an通项公式为
an=a1+(n-1)d=a1+(n-1)a1/2=(n+1)a1/2
a5/a1=3
所以kn+1是公比为3的等比数列
kn+1=2*3^n
kn=2*3^n-1,n=0,1.
k1+k2+k3+.+kn
=2*3^0-1+2*3^1-1+.
=2(3^0+3^1+3^2+.)-n
=2(3^n-1)/2-n
=3^n-1-n
a1,a5,a17为等比数列
(a5)^2=a1*a17
(a1+4d)^2=a1(a1+16d)
16d^2-8a1d=0
a1=2d
an通项公式为
an=a1+(n-1)d=a1+(n-1)a1/2=(n+1)a1/2
a5/a1=3
所以kn+1是公比为3的等比数列
kn+1=2*3^n
kn=2*3^n-1,n=0,1.
k1+k2+k3+.+kn
=2*3^0-1+2*3^1-1+.
=2(3^0+3^1+3^2+.)-n
=2(3^n-1)/2-n
=3^n-1-n