1.設a1=a
則1/a1a2 + 1/a2a3 + .+ 1/a(n-1)an
=1/a(a+d)+1/(a+d)(a+2d)+……+1/[a+(n-2)d][a+(n-1)d]
={d/a(a+d)+d/(a+d)(a+2d)+……+d/[a+(n-2)d][a+(n-1)d]}/d
={1/a-1/(a+d)+1/(a+d)-1/(a+2d)+……+1/[a+(n-2)d]-1/[a+(n-1)d]}/d
={1/a-1/[a+(n-1)d]}/d
=[1/a-1/(a+nd-d)]/d
=(a+nd-d-a)/d(a+nd-d)
=(nd-d)/d(a+nd-d)
=(n-1)/(a+nd-d)
2.
(1) a1+3a2+3^2a3+...+3^(n-1)an=n/3
a1+3a2+3^2a3+...+3^(n-2)a(n-1)=(n-1)/3
2式相減
3^(n-1)an=1/3
an=1/3^n
(2 ) bn=n/an=n*3^n
Sn=3+2*3^2+3*3^3+.+n*3^n
3Sn=3^2+2*3^3+3*3^4...+(n-1)*3^n+n*3^(n+1)
2式相減
-2Sn=3+3^2+3^3+.+3^n-n*3^(n+1)
-2Sn=3(1-3^n)/(1-3)-n*3^(n+1)
-2Sn=-[3-3^(n+1)]/2-n*3^(n+1)
Sn=[3-3^(n+1)]/4+n*3^(n+1)/2
3.
(1)、
因為an+1=2Sn
所以2an=2(Sn-Sn-1)=2Sn-2Sn-1=2an+1+2-2an-2=2an+1-2an
兩邊加2an得4an=2an+1既an+1=2an
因此{an}為首項1公比2的等比數(shù)列
an=2^(n-1)
(2)、
因為nan=n*2^(n-1)
所以Tn=1*2^0+2*2^1+3*2^2+4*2^3+...+n*2^(n-1)
等號兩邊乘2
2Tn=1*2^1+2*2^2+3*2^3+4*2^4+...+n*2^n
所以Tn=2Tn-Tn
=[1*2^1+2*2^2+3*2^3+4*2^4+...+n*2^n]
-[1*2^0+2*2^1+3*2^2+4*2^3+...+n*2^(n-1)]
=-(2^0+2^1+2^2+2^3+2^4+...+2^(n-1)]+n*2^n
=-(2^n-1)+n*2^n
=1+(n-1)*2^n