Trivial Question : generating function

Question

How does $ 1 + x + x^2 + x^3 + x^4 = \frac {1-x^5}{1-x} $ ?

Answer 1

If you know that $\dfrac{1}{1-x} = 1 + x + x^2 + \ldots$, then you can use this fact after breaking up the fraction:

$$ \begin{align} \frac{1-x^5}{1-x} &= \frac{1}{1-x} - \frac{x^5}{1-x} \\ &= \frac{1}{1-x} - x^5\frac{1}{1-x} \\ &= (1 + x + x^2 + \ldots) - x^5(1 + x + x^2 + \ldots) \\ & = 1 + x + x^2 + x^3 + x^4 \end{align} $$

This is effectively the reverse direction of some of the other answers.

Answer 2

$(1 + x + x^2 + x^3 + x^4)(1-x)=(1 + x + x^2 + x^3 + x^4) - (x + x^2 + x^3 + x^4+x^5) = 1-x^5$

Answer 3

$\bf{My\; Solution::}$ Let $$\displaystyle S = 1+x+x^2+x^3+x^4\tag1$$

Now Multiply both side by $x\;,$

We Get $$x\cdot S = x+x^2+x^3+x^4+x^5\tag2$$

Now $(1)-(2)\;,$ We Get $$\displaystyle (1-x)\cdot S = 1-x^5\Rightarrow S = \frac{1-x^5}{1-x}$$

Answer 4

Think of $1+x+x^2+x^3+x^4$ as a geometric series with common ratio $x$. The sum of the first $n$ terms of a geometric series is given by $$S_n = \frac{a(1-r^n)}{1-r}$$where $a$ is the first term and $r$ the common ratio. Here $a = 1$, $r = x$, $n = 5$ so we get $$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$

Answer 5

All this comes from the high school identity: $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1})$$ and setting $a=1$.

Answer 6

Let $S= 1 + x + x^2 + x^3 + x^4 \quad I$

multiplying both sides by x

$xS=x + x^2 + x^3 + x^4+x^5 \quad II$

Subtracting II by I

$S(1-x)=1-x^5$

Dividing the equation by $(1-x)$

$S=\frac{1-x^5}{1-x}$