（1）容易觀察到1,4,...,3n-2是公差為3的等差數(shù)列；
1/2,1/4,...,1/2^n是公差為1/2的等比數(shù)列.
Sn=1又1/2+4又1/4+7又1/8+…+[(3n-2)+1/2^n]
=[1+4+7+...+(3n-2)]+(1/2+1/^2+...+1/2^n)
=(3n-2+1)n/2+(1/2)(1-1/2^n)/(1-1/2)
=（3n-1)n/2+1-2(-n)
(2)可觀察到各項系數(shù)是公差為3的等差數(shù)列,x,x^2,...,x^n是公比為x的等比數(shù)列
當x=0時,Sn=0；
x=1時,Sn=1+4+...+(3n-2)=(3n-2+1)n/2=(3n-1)n/2；
當x≠0和1時:
Sn=x+4x^2+7x^3+…+(3n-2)x^n …①
①式兩邊同時乘以x得：
xSn=x^2+4x^3+7x^4+…+(3n-2)x^(n+1)…②
②-①得：
（x-1)Sn=-x-(3x^2+3x^4+...+3x^n)+(3n-2)x^(n+1)
=-x-3(x^2+x^3+...+x^n)+(3n-2)x^(n+1)
=-x-3x^2[x^(n-1）-1]/（x-1)+(3n-2)x^(n+1)
故：Sn=-x/(x-1)-3x^2[x^(n-1）-1]/（x-1)^2+(3n-2)x^(n+1)/(x-1)
結果不用化簡
(3）令Sn=1/a1a2+1/a2a3+…+1/an*a(n+1)
則Sn=1/2*5+1/5*8+...+1/(3n-1)(3n+2)
=(1/3)(1/2-1/5)+(1/3)（1/5-1/8)+...+(1/3)[1/(3n-1)-1/(3n+2)]
=(1/3)[1/2-1/5+1/5-1/8+...+1/(3n-1)-1/(3n+2)](觀察到中間各項約去,只剩首末兩項）
=(1/3)[1/2-1/(3n+2)]
=n/[2(3n+2)]