sn=1/x+2/x^2+3/x^3+.n/x^n求sn
sn=1/x+2/x^2+3/x^3+.n/x^n求sn
Sn=1/x+2/x^2+3/x^3+...+n/x^n
Sn/x=1/x^2+2/x^3+..+n/x^(n+1)
Sn-[Sn/x-n/x^(n+1)]=(1/x+1/x^2+1/x^3+..+1/x^n)
(1-1/x)Sn=(1/x)[1-(1/x)^(n+1)]/(1-1/x)+n/x^(n+1)
Sn=x*[1-(1/x)^(n+1)]/(x-1)^2+n/[x^(n-1)*(x-1)^2]
求导y=1/x+2/x^2+3/x^3
解 y=x^（-1）+2x^（-2）+3x^（-3）
y‘=-x^（-2）-4x^(-3）-9x^（-4)
Sn=1/(1x2x3)+1/(2x3x4)+.+1/(n(n+1)(n+2))
解.裂项法.
Sn=1/[n(n+1)(n+2)]=(1/2){1/[n)n+1)]-1/[(n+1)(n+2)]}
=(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]
=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/n(n+1)(n+2)
=(1/2)[1/1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+/4-2/5+1/6
+.....+1/n-2/(n+1)+1/(n+2)]
=(1/2)[1-1/2-1/(n+1)+1/(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
其他类似的题目比如：
1/1x2x3+1/2x3x4+1/3x4x5+------+1/98x99x100
=(1/2)*(1/1*2-1/2*3)+(1/2)*(1/2*3-1/3*4)+...+(1/2)(1/98*99-1/99*100)
=(1/2)*(1/1*2-1/2*2+1/2*3-1/3*4+...+1/98*99-1/99*100)
=(1/2)*(1/2-1/9900)
=(1/2)*(4949/9900)
=4949/19800.
求Sn=1/1x2+1/2x3+1/3x4+.+1/n(n+1)
裂项法：
Sn=1/1x2+1/2x3+1/3x4+....+1/n(n+1)
=1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
设函数f(x)=2^x(2^x+根号2),若Sn=f(1)+f(1/2)+f(1/3)+.+f(1/n),求Sn
2的X次方，应该是X的平方吧
，不然Sn无穷大。
f(1)=1+根号2，f(1/2)=1/4（1/4+根号2），
得出，Sn=1+1/（2）4次方+1/（3）4次方+……+1/（n）4次方+ 根号2（1+1/2的平方+1/3的平方+……+1/n平方）
sn=x/1+x^2/2!+x^3/3!+.+x^n/n!怎么求和？
解：因为e^x=1+x/1!+x^2/2!+x^3/3!+…+x^n/n!+…，故
当n->+∞时，才有
sn= x/1+x^2/2!+x^3/3!+...+x^n/n!=e^x-1
n->+∞ n->+∞
Sn=1/（1x3）+1/（3x5）+…1/（（2n-1）（2n+1））
典型的裂项相消法
具体步骤如下：Sn=1\2(1-1\3+1\3-1\5+1\5-1\7...1\(2n-1)-1\(2n+1) )
=1\2(1-1\(2n+1) )
=1\2*[2n\(2n+1）】
=n\(2n+1)
祝你学习进步！学业有成！
求Sn=1x(1/2)^0+2x(1/2)^1+3x(1/2)^2+---+nx(1/2)^n-1
Sn=1x(1/2)^0+2x(1/2)^1+3x(1/2)^2 +---+nx(1/2)^n-1
1/2Sn= 1x(1/2)^1+2x(1/2)^2+---+(n-1)x(1/2)^n-1+nx(1/2)^n
sn-1/2sn=1/2sn=1+(1/2)^1+(1/2)^2+---+(1/2)^(n-1)-nx(1/2)^n
=[1-(1/2)^n]/(1-1/2)-nx(1/2)^n
sn=1-(1/2)^n-nx(1/2)^(n-1)
已知函数f(x)=1/2x^2+3/2x,数列an的前n项和为Sn,点(n,Sn)[n属于N*]均在
1. =(1/2)n^2+(3/2)n-(1/2)(n-1)^2-(3/2)(n-1)=n+1n=1时，A1=S1=1/2+3/2=2也满足上式2. Bn=An/2^(n-1)=(n+1)/2^(n-1) Tn=B1+B2+B3+……+Bn =2/2^0+3/2^1+4/2^2+……+(n+1)/2^(n-1)两边同乘22Tn=4+3/2^0+4/2^1+……+(n+1)/2^(n-2) 两式错位相减 2Tn-Tn=4+[(3/1-2/1)+(4/2-3/2)+……+(n+1)/2^(n-2)-n/2^(n-2)]-(n+1)/2^(n-1) =4+(1+1/2+……+1/2^(n-2)-(n+1)/2^(n-1) =4+(1-1/2^(n-1))/(1-1/2)-(n+1)/2^(n-1) =6-(n+3)/2^(n-1)3. Cn=An/A(n+1)+A(n+1)/An=(n+1)/(n+2)+(n+2)/(n+1) Cn-2=(n+1)/(n+2)+(n+2)/(n+1)-2 =(n+2-1)/(n+2)+(n+1+1)/(n+1)-2 =1-1/(n+2)+1+1/(n+1)-2 =1/(n+1)-1/(n+2) C1+C2+……+Cn-2n =(1/2-1/3)+(1/3-1/4)+……+1/(n+1)-1/(n+2) =1/2-1/(n+2)<1/2 C1+C2+……+Cn-2n<1/2 C1+C2+……+Cn<2n+1/2 设a=(n+1)/(n+2) b=(n+2)/(n+1) 则ab=1 a+b>=2(ab)^0.5 当且仅当a=b时，不等式取等号 Cn=a+b>=2(ab)^0.5=2 若a=b 则(n+1)/(n+2)=(n+2)/(n+1) 解得n=-3/2，不属于N 不等式等号不成立Cn>2C1+C2+……+Cn>2n综上所述2n<C1+C2+……+Cn<2n+1/2
Sn＝（x＋1/x）^2＋（x^2＋1/x^2）^2.＋（x^n＋1/x^n）^2求和
Sn =x^2+1/x^n+2+x^4+1/x^4+2+.........+x^(2n)+1/x^(2n)+2
=[x^2+x^4+.......+x^(2n)]+[1/x^2+1/x^4+.........+1/x^(2n)]+2n
x≠±1时，Sn=x²(x^2n-1)/(x²-1)+(1-1/x^(2n))/(x²-1)+2n
x=±1时，Sn=4n
