错位相减法数列求和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;这一步看不懂,
人气:372 ℃ 时间:2019-10-26 05:31:10
解答
Sn=1+3+5+…+(2n-1)
Sn=(2n-1)+(2n-3)+(2n-5)+…+1
2Sn=(1+2n-1)+(3+2n-3)+(5+2n-5)+…+(2n-1+1)=2n+2n+2n+…+2n=2n²
Sn=n²
推荐
- 数列求和 Sn=x+2x^2+3x^3+…+nx^n求详细过程错位相减法算的
- 用错位相减法求Sn=1+3x+6x^2+9x^3+……+(2n-1)x^(n+1).
- 求数列3x,5x^2,7x^3,...,(2n+1)x^n,...的前n项和
- 如何用错位相减法算求数列1乘2,3乘2^2,5乘2^3,...,(2n-1)乘2^n的前n项和Sn
- 数列求和 用错位相减法求和 1/a+2/a^2+3/a^3+…+n/a^n (a不等于0)
- Your father and mother who plays more important roles in your daily life
- 已知多项式A=x^2-xy+y^2;B=x^2+xy+y^2 (1) 求2A-2B
- 甲乙两人同时从AB两地相向而行,甲行6小时与乙在距中点12千米相遇,这时乙行了42千米甲每小时行多少千米?
猜你喜欢