求前N项和 Sn=1又1/2+4又1/4+7又1/8+……[(3n-2)+1/2^n]
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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)]
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)]
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)]
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