Sn=x+3x+5x^2+7x^3+…+(2n-1)*x^(n-1)(x≠0)
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/01 08:00:44
x)γ66363~L[(OPS"NDkT<\`iTOz;ҭyrmygËuXdǮ'ڞʼXt_:/f=jӽl
MߦW|V5J
u5/.H 5r
求和:Sn=1+3x+5x*x+7x*x*x+……+(2n-1)x^n-1 (x不为0和1)
求和Sn=(x-1)+(x^3-2)+(x^5-3)+(x^7-4)+…+(x^2n-1-n)
求和:Sn=1+3x+5x+7x+…+﹝2n-1﹞x^n-1
Sn=x+3x+5x^2+7x^3+…+(2n-1)*x^(n-1)(x≠0)
求和:Sn=x+2x*x+3x*x*x+.+nXn
求和:Sn=x+2x^2+3x^3+……+nx^x
求和Sn=(x-1)+(x^2-2)+(x^3-3)+…+(x^n-n)
求Sn=x+2x平方+3x平方…nxn次方(x不等于0)
sn=(x-1)+(x^2-3)+(x^3-5)+...+[x^n-(2n-1)]
求Sn=3x+4x²+5x³+.+(n+2)x^n
错位相减法的题如:求和Sn=1+3x+5x^2+7x^3+…+(2n-1)*x^(n-1)(x≠0)当x=1时,Sn=1+3+5+…+(2n-1)=n^2;当x不等于1时,Sn=1+3x+5x^2+7x^3+…+(2n-1)*x^(n-1);∴xSn=x+3x^2+5x^3+7x^4+…+(2n-1)*x^n;两式相减得(1-x)Sn=1+2[x+x^2+x^3
sn=1*x+3*x^2+5*x^3……+(2n-1)*x^n.求和
求和Sn=1+2x+3x^2+4x^3+5x^4……+nx^n-1
求和:Sn=1-3x+5x^2-7x^3+.+(2n+1)(-x)^n(n属于N*)
求和:Sn=1+3x+5x+7x+...+(2n-1)x^(n-1)谢谢了,
错位相减法的问题,例如,求和Sn=x+3x+5x^2+7x^3+…+(2n-1)*x^(n-1)(x≠0)当x=1时,Sn=1+3+5+…+(2n-1)=n^2;当x不等于1时,Sn=1+3x+5x^2+7x^3+…+(2n-1)*x^(n-1);∴xSn=x+3x^2+5x^3+7x^4+…+(2n-1)*x^n;两式相减得(1-x)Sn=1+2x[1
*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X ||
Sn=1+3x+5x^2+7x^3+…+(2n-1)*x^(n-1); ∴xSn=x+3x^2+5x^3+7x^4+…+(2n-1)*x^n; 两式相减得(1-x)Sn=1+2x[1+x+x^2+x^3+…+x^(n-2)]-(2n-1)*x^n(两式相减后的这个结果是如何得到的?)