Let us consider what happens when we try to expand $$ (x+h)^n = (x+h)(x+h)\cdots (x+h) .$$ To expand the RHS, we must pick either $x$ or $h$ from each bracket, and multiply them …

Harmonic numbers are related to the harmonic mean in that the n-th harmonic number is also n times the reciprocal of the harmonic mean of the first n positive integers. Harmonic numbers …

2017年2月28日 · 若x和h均为整数，则可以说h是 (x + h)^n - x^n的一个因子或 (x + h)^n - x^n可被h整除。 求数学大师，（x+h）ⁿ-xⁿ即x和h之和的n次方减去x的n次方，怎么展开，谢谢可利用 …

2014年7月13日 · 是二项式定理的展开式把，(x+h)^n=Cn^0*x^n+Cn^1*x^n-1h^1+…+Cn^r*x^n-rh^r+…+Cn^n*h^n(n∈N*), 希望可以帮到你

2023年1月11日 · When you expand $(x+h)^n$ there will be $2^n$ terms. Each will have some $x$ 's and some $h$ 's. Precisely $n$ of them will have one $h$ and $n-1$ $x$ 's. The …

2015年6月23日 · I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative: \begin{align} (x^n)'&=\lim_{h \to 0} {(x+h)^n-x^n\over h}\\ &=\lim_{h \to 0} {x^n+nx^{n-1}h+{n(n-1)\over 2}x^{n …

把（x+h）^n用 二项式定理 展开。然后第一项可以和x^n约掉。剩下的把h=o带入。就剩下nx^（n-1）

令x表示任意实数，h表示一个小的增量，n表示对应的n次方。 求（x+h）的n次方展开式可以写作： x+h的n次方展开式-求 (x+h)的n次方展开式是我们在学习微积分课程中必不可少的技巧，它 …

Harmonic numbers appear in many expressions involving special functions in analytic number theory, including the Riemann zeta function. They are also related closely to the natural …

We would like to expand and simplify this expression, but \((x+h)^n\) is not so simple to expand when we don’t have a specific value for \(n\). However, \(x+h\) is a binomial. We have a way to …