an=1/（3n-2) 记Sn=a1*a2+a2*a3+...+an*a(n+1) 求证：Sn＜1/3
问题描述：
an=1/（3n-2) 记Sn=a1*a2+a2*a3+...+an*a(n+1) 求证：Sn＜1/3
an=[2的（n-2)次方]乘以（3n-1）求前n项和公式
答
1.an*an+1=1/[（3n-2)(3n+1)]=1/3[1/(3n-2)-1/(3n+1)]
Sn=1/3[1-1/4+1/4-1/7+…+1/(3n-2)-1/(3n+1)]=1/3[1-1/(3n+1)]＜1/3
2.an=(3n-1)*2^(n-2)
Sn=2(1/2)+5+8*2+…+(3n-1)*2^(n-2) ①
2Sn=2*[2(1/2)+5+8*2+…+(3n-1)*2^(n-2)]
=2+5*2+8*4+…+(3n-4)*2^(n-2)+(3n-1)*2^(n-1) ②
②-①:Sn=(3n-1)*2^(n-1)-1-3*[1+2+2^2+…+2^(n-2)]
=(3n-1)*2^(n-1)-1-3*[2^(n-1)-1]
=(3n-4)*2^(n-1)+2.