How do I prove that $$\dfrac{1-x^{11}}{1-x}=1+x+x^2+x^3+\cdots+x^{10} $$
Attempt: By observing first let assume that $∣x∣<1$ then $\dfrac{1-x^{11}}{1-x}=\dfrac{1}{1-x}(1-x^{11})=(1+x+x^2+x^3+x^4+\cdots +x^n +\cdots)(1-x^{11})$
Any idea or hint?
Asked
Active
Viewed 68 times
0
-
1See https://en.wikipedia.org/wiki/Geometric_progression#Geometric_series – lab bhattacharjee Apr 22 '17 at 09:33
-
2The easiest way is to approve that $(1-x)(1+x+x^2+x^3+\cdots+x^{10})$ is equal to $1-x^{11}$ by just calculating the product. Nearly all terms will cancel out. Then, the equation must hold for $x\ne 1$ – Peter Apr 22 '17 at 09:36
-
Gotcha. Thanks. – Fawad Apr 22 '17 at 09:38
-
the equation $$\dfrac{1}{1-x}(1-x^{11})=(1+x+x^2+x^3+x^4+\cdots +x^n +\cdots)(1-x^{11})$$ is not true for $|x|> 1$ so it is better to just make the direct product of $1-x$ and $1+x+\ldots+x^{10}$ as @Peter suggested in the second commentary, noticing that $\frac{1-x^{11}}{1-x}$ is just defined for $x\neq 1$. – Masacroso Apr 22 '17 at 09:52
-
1This is a special case of Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$. (Perhaps they are close enough to be considered duplicates?) – Martin Sleziak Apr 22 '17 at 10:21
-
hint: $1-x^{11}= 1+x-x+x^2-x^2+x^3-x^3+x^4-x^4+x^5-x^5+x^6-x^6+x^7-x^7+x^8-x^8+x^9-x^9+x^{10}-x^{10}-x^{11}$ – John Joy Apr 26 '17 at 14:26