If $n$ is a positive integer and $x$ is any non-zero real number, then prove that $c_0+c_1\dfrac{x}{2}+c_2\cdot\dfrac{x^2}{3}+c_3\cdot\dfrac{x^3}{4}+\ldots+c_n\cdot\dfrac{x^n}{n+1}=\dfrac{(1+x)^{n+1}-1}{(n+1)x}$.

Question

Accepted Answer

>

> [!TLDR]
> Start from the binomial expansion $(1+t)^n=\sum c_r t^r$, integrate both sides with respect to $t$ from $0$ to $x$, then divide by $x$.
>
> By the binomial theorem,
> $$(1+t)^n = c_0+c_1t+c_2t^2+c_3t^3+\cdots+c_nt^n \quad (\star)$$
> where $c_r={}^nC_r$.
>
> **Integrate both sides** with respect to $t$ from $0$ to $x$ (valid for any real $x
e0$):
>
> Left side:
> $$\int_0^x(1+t)^n\,dt = \left[\frac{(1+t)^{n+1}}{n+1}ight]_0^x = \frac{(1+x)^{n+1}-1}{n+1}$$
>
> Right side: integrate term by term,
> $$\int_0^x\left(c_0+c_1t+c_2t^2+\cdots+c_nt^night)dt = c_0x+c_1\frac{x^2}{2}+c_2\frac{x^3}{3}+\cdots+c_n\frac{x^{n+1}}{n+1}$$
>
> Equating the two sides:
> $$\frac{(1+x)^{n+1}-1}{n+1} = c_0x+c_1\frac{x^2}{2}+c_2\frac{x^3}{3}+\cdots+c_n\frac{x^{n+1}}{n+1}$$
> …

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Q.If $n$ is a positive integer and $x$ is any non-zero real number, then prove that $c_0+c_1\dfrac{x}{2}+c_2\cdot\dfrac{x^2}{3}+c_3\cdot\dfrac{x^3}{4}+\ldots+c_n\cdot\dfrac{x^n}{n+1}=\dfrac{(1+x)^{n+1}-1}{(n+1)x}$ .

Concept understanding — Binomial Theorem

The Binomial Theorem: From Patterns to Power

The Pattern You Already Know

Why Do These Coefficients Appear?

The Precise Statement

A Quick Example

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