-1

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

lapin
  • 459

6 Answers6

3

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

JMP
  • 21,771
3

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

Rick
  • 2,926
2

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.

Théophile
  • 24,627
  • 1
    How do we convert a power series to a polynomial as in this case from $ 1+x+x^2+... to; 1 + x+x^2+x^3+x^4 ? $ i.e. How do we know to multiply by $ (1-x^5) $ ? – lapin Aug 07 '15 at 13:13
  • 1
    Starting from $S = 1+x+x^2+\ldots$, if you want to transform it into $1+x+x^2+x^3+x^4$, then look at what needs to be removed, which is $x^5+x^6+x^7+\ldots$; let's call this series $T$. So far, we have $1+x+x^2+x^3+x^4 = S-T$. Next it's a matter of simplifying $T$: taking out the common factor, $x^5$, we have $$T = x^5(1+x+x^2+\ldots) = x^5S.$$ Therefore $1+x+x^2+x^3+x^4 = S - x^5S = (1-x^5)S$. – Théophile Aug 07 '15 at 13:27
1

$\bf{My\; Solution::}$ 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 $$\displaystyle (1-x)\cdot S = 1-x^5\Rightarrow S = \frac{1-x^5}{1-x}$$

Théophile
  • 24,627
juantheron
  • 53,015
1

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 $a=1$.

Bernard
  • 175,478
1

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 $(1-x)$

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

callculus42
  • 30,550