作业帮求前N项和 Sn=1又1/2+4又1/4+7又1/8+……[(3n-2)+1/2^n]Sn=x+4x^2+7x^3+……+(3n-2)x^n已知an=3n-1 求(1/a1a2)+(1/a2a3)+……[1/an*A(n+1)]
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![求前N项和 Sn=1又1/2+4又1/4+7又1/8+……[(3n-2)+1/2^n]Sn=x+4x^2+7x^3+……+(3n-2)x^n已知an=3n-1 求(1/a1a2)+(1/a2a3)+……[1/an*A(n+1)]](/uploads/image/z/1389337-25-7.jpg?t=%E6%B1%82%E5%89%8DN%E9%A1%B9%E5%92%8C+Sn%3D1%E5%8F%881%2F2%2B4%E5%8F%881%2F4%2B7%E5%8F%881%2F8%2B%E2%80%A6%E2%80%A6%5B%283n-2%29%2B1%2F2%5En%5DSn%3Dx%2B4x%5E2%2B7x%5E3%2B%E2%80%A6%E2%80%A6%2B%283n-2%29x%5En%E5%B7%B2%E7%9F%A5an%3D3n-1+%E6%B1%82%281%2Fa1a2%29%2B%281%2Fa2a3%29%2B%E2%80%A6%E2%80%A6%5B1%2Fan%2AA%28n%2B1%29%5D)
求前N项和 Sn=1又1/2+4又1/4+7又1/8+……[(3n-2)+1/2^n]Sn=x+4x^2+7x^3+……+(3n-2)x^n已知an=3n-1 求(1/a1a2)+(1/a2a3)+……[1/an*A(n+1)]
求前N项和 Sn=1又1/2+4又1/4+7又1/8+……[(3n-2)+1/2^n]
Sn=x+4x^2+7x^3+……+(3n-2)x^n
已知an=3n-1 求(1/a1a2)+(1/a2a3)+……[1/an*A(n+1)]
求前N项和 Sn=1又1/2+4又1/4+7又1/8+……[(3n-2)+1/2^n]Sn=x+4x^2+7x^3+……+(3n-2)x^n已知an=3n-1 求(1/a1a2)+(1/a2a3)+……[1/an*A(n+1)]
（1）容易观察到1,4,...,3n-2是公差为3的等差数列；
1/2,1/4,...,1/2^n是公差为1/2的等比数列.
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)可观察到各项系数是公差为3的等差数列,x,x^2,...,x^n是公比为x的等比数列
当x=0时,Sn=0；
x=1时,Sn=1+4+...+(3n-2)=(3n-2+1)n/2=(3n-1)n/2；
当x≠0和1时:
Sn=x+4x^2+7x^3+…+(3n-2)x^n …①
①式两边同时乘以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)
结果不用化简
(3）令Sn=1/a1a2+1/a2a3+…+1/an*a(n+1)
则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)](观察到中间各项约去,只剩首末两项）
=(1/3)[1/2-1/(3n+2)]
=n/[2(3n+2)]