Formula for expanding powers

Question

I can't seem to understand the following: $$A^n-B^n=\left(A-B\right)\cdot \left(A^{n-1}+A^{n-2}\cdot B+\cdots+A\cdot B^{n-2}+B^{n-1}\right)$$

How can i derive this formula ? Also, when does the $A^\text{something}$ end and when does $B^\text{something}$ start.

Thank you.

Hint. What does this equality say when $n=2$? The second factor on the right then has just two terms. Now look at $n=3$. The second factor will have three terms. — Ethan Bolker, Sep 18 '17 at 13:21
Note that the powers of A and B don't start or end - they are present in every term. We begin with $A^{n-1}\cdot B^0$, then the next term is found by dividing by A and multiplying by B (i.e. $A^{n-2}\cdot B^1$, $A^{n-3}\cdot B^2$, ...), until we reach $B^{n-1}$. — A.M., Sep 18 '17 at 13:24
It might help to think carefully about the distributive property of multiplication over addition: $(A-B) C = AB - AC$. Let $C$ be the sum, then distribute both $A$ and $B$ across the sums (this requires two distribution steps). You should see a pattern. — Xander Henderson, Sep 18 '17 at 13:30
Possible duplicate of Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$. Also related. — Xander Henderson, Sep 18 '17 at 13:34

score 1 · Answer 1 · edited Sep 18 '17 at 13:53

1

Hint: Write it in sum notation and try to work out the brackets. $$ \left(A-B\right)\cdot \left(A^{n-1}+A^{n-2}\cdot B+\cdots+A\cdot B^{n-2}+B^{n-1}\right) = (A-B)\sum_{i=0}^{n-1}A^{n-i-1}B^{i} = \cdots $$

edited Sep 18 '17 at 13:53

Michael Hardy

1

answered Sep 18 '17 at 13:21

Marc

6,861

For me at least, "writing in sum notation" makes the problem harder, not easier. When I see sum notation I convert it to a form with dot-dot-dots. Perhaps best to just "work out the brackets" (that is, distribute) right from the start. – Ethan Bolker Sep 18 '17 at 13:57
I find both useful when solving a problem, but the sum notation seems more formal and rigorous to me. Who is to say that the pattern in $A^{n-1}+A^{n-2}\cdot B+\cdots$ does not continue as something like this: $A^{n-1}+A^{n-2}\cdot B+A^{n-4}\cdot B^2 + A^{n-8}\cdot B^3+\cdots+A^{n-2^k}\cdot B^k+\cdots$? – A.M. Sep 18 '17 at 17:48

Marios Gretsas · Answer 2 · 2017-09-18T14:56:02.503

You can do the computations:

$$A\left(A^{n-1}+A^{n-2}\cdot B+\cdots+A\cdot B^{n-2}+B^{n-1}\right)$$ $$=A^n+A^{n-1}B+A^{n-2}B^2+....+A^2B^{n-2}+AB^{n-1}$$

.

$$B\left(A^{n-1}+A^{n-2}\cdot B+\cdots+A\cdot B^{n-2}+B^{n-1}\right)$$ $$=BA^{n-1}B^2A^{n-2}+A^{n-3}B^3+....+B^n$$

If substract: $$A\left(A^{n-1}+A^{n-2}\cdot B+\cdots+A\cdot B^{n-2}+B^{n-1}\right)-B\left(A^{n-1}+A^{n-2}\cdot B+\cdots+A\cdot B^{n-2}+B^{n-1}\right)=A^n-B^n$$

In other words,if you notice, everything will be canceled out and the only thing remaining will be $A^n-B^n$

score 1 · Answer 3 · answered Sep 18 '17 at 17:40

Note: $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots+AB^{n-2}+B^{n-1}),$$ where: $$S_n=A^{n-1}+A^{n-2}B+\cdots+AB^{n-2}+B^{n-1}$$ is the sum of the first $n$ terms of geometric progression with the first term $A^{n-1}$ and the common ratio $\frac{B}{A}$, hence: $$(A-B)\cdot S_n=(A-B)\cdot \frac{A^{n-1}\left(1-\left(\frac{B}{A}\right)^n\right)}{1-\frac{B}{A}}=A^n-B^n.$$

score 0 · Answer 4 · answered Sep 18 '17 at 14:44

when does the $A^\text{something}$ end and when does $B^\text{something}$ start

$$ A^6 - B^6 = (A-B)(A^5 + A^4B + A^3B^2 + A^2B^3+AB^4+B^5) $$ When $n=6$ then the powers of $A$ are $5,4,3,2,1,0$ in that order, and the powers of $B$ are $0,1,2,3,4,5$ in that order.

Formula for expanding powers

4 Answers4