1

I'm a little stuck on proving $$(b^n-a^n)=(b-a)(a^0b^{n-1}+a^1b^{n-1}+ \cdots + a^{n-1}b^{0}).$$A solution I came across gave this as an answer: $$(b-a)(b^n + b^{n-1} a+...+ba^{n-1} +a^{n} )\\ =(b-a)b^n + (b-a)b^{n-1} a+...+(b-a)ba^{n-1} +(b-a)a^n\\ =b^n+1 -b^n a+b^n a-b^{n-1} a^2 +...+b^2 a^{n-1} -ba^n +ba^n -a^{n+1}\\ =b^{n+1} -a^{n+1}$$

Since for each $i=0,\ldots,n$, notice that for each term $b^{n-i} a^{i}$ there is a $-b^{n-i} a^{i}$, so everything cancels except for $b^{n+1} -a^{n+1}$ , so $$(b-a)(b^n + b^{n-1} a+\cdots +ba^{n-1} +a^n )= b^{n+1} - a^{n+1}$$

But how does proving that $b^{n+1} - a^{n+1}$ is true prove the formula?

Arthur
  • 199,419
Nikitau
  • 1,353
  • Is your question "This proves the formula for $b^{n+1} - a^{n+1}$, but how do you prove it for $b^n - a^n$?" – Arthur Sep 23 '16 at 04:40
  • No, my question is how does proving it for n+1 prove it for the n case? I came across the solution online and I was thoroughly confused. – Nikitau Sep 23 '16 at 04:44
  • 1
    Do you accept the argument you quote for proving the formula for $b^{n+1}-a^{n+1}$? It is reasonably convincing, but the dot dot dot in the middle leaves a bit of room for doubt. If you do, set $n+1=k$ and you have proven it for $a^k-b^k$. If not, you need to formally do the proof using summation signs to make sure you get all the terms. It will come out OK. – Ross Millikan Sep 23 '16 at 04:46
  • @RossMillikan I accept the argument and I understand it. I guess I could expand the .. further? If I leave it as n+1, then is the proof still satisfied? Could I let n=n+1 since n is just any arbitrary number? – Nikitau Sep 23 '16 at 04:48
  • n can be any variable. So you can just relabel it as n+1. If something is true for any n+1 then it is true for n and it is true for n+7. If you don't like it preceded it with "let m=n-1" then prove it for "m+1" and point out "m+1=n" so it is proven for "n". Or Stat the first line with $(b-a)(b^{n-1} + b^{n-2}a + .... + ba^{n-2} + a^{n-1}) = (b^n-a^n) $ – fleablood Sep 23 '16 at 04:56
  • The only reason you might have trouble with $n+1$ instead of $n$ is if it isn't true for $1$ and they are letting $n$ range over $1,2,3,\ldots $. In this case there is no problem as $b^1-a^1=b-a$ is your base case. It would have been clearer to prove it for $b^n-a^n$, which works the same way and would avoid your confusion. – Ross Millikan Sep 23 '16 at 05:05
  • Thanks for clarifying! Unfortunately I couldn't find an alternate proof that gives back $b^n - a^n$. I did try proof by induction, but I ended up getting stuck. – Nikitau Sep 23 '16 at 05:11
  • It is odd that the showed it is true for n+1 rather than n. But if you simply "reset" n to n+1 it is no problem. There is the assumption that it is true for all $n >= 1$ so it is true for all $n+1$ where $n >= 0$. But once, you've proven for all $n$ you've prove it for $7$ when $n = 6$ and you've proven it for $8$ when $n=7$. So if you've proven it for $7$, does it really make any difference if you proved it for $n+1=7$ or $n=7$? As long as it is true for $7$ does it matter if $7$ is $n $ or $k $ or $n+1$. It's true for all values. n, or n+1, is just a freakin' variable. – fleablood Sep 23 '16 at 05:21
  • @fleablood Yes, I understand the concept -- I've already mentioned a few comments up on letting n = n+1 since n itself is arbitrary. – Nikitau Sep 23 '16 at 05:42

2 Answers2

0

If it's true for $n+1$ then it is true for $m=n+1$. And if is true for $m $ then it is true for $n=m $.

Or, more generally, if you prove it for any arbitrary value, you have proven it for all arbitrary values. $n$ (for $n \ge 1$) and $n+ 1$ (for $n \ge 0$) or $k +357$ for ($k \ge -356$) are equally arbitrary.

$(b-a)(b^{k+356}+ b^{k+355}a+...+ba^{k+355}+ a^{k+356})=$

$b^{k+357}-a^{357}) $

Is it not clear we have proven it for all variables?

Wha if we were asked to prove $x-y $ divides $x^h - y^h $? Would we have to prove it all over again just because we proved it for $a,b,n+1$ but not for $y,x,h $?

===

"Unfortunately I had trouble coming up with an alternative proof that gave back $b^n -a^n$"

Seriously???

$(b-a)(a^0b^{n-1} + a^1b^{n-2}+...+a^{n-2}b^1+a^{n-1}b^0)=$

$b ( (a^0b^{n-1} + a^1b^{n-2}+...+a^{n-2}b^1+a^{n-1}b^0) - a (a^0b^{n-1} + a^1b^{n-2}+...+a^{n-2}b^1+a^{n-1}b^0)=$

$b^n +(ab^{n-1}-ab^{n-1}) +....+(a^{n-1}b-a^{n-1}b) - a^n=$

$b^n-a^n $

You weren't able to modify the proof to do it for one lower power?

fleablood
  • 124,253
0

To prove that $$\begin{align} b^n-a^n &= (b-a)(a^0b^{n-1}+a^1b^{n-1} + \cdots + a^{n-1}b^{0})\\ &=(b-a)\sum_{k=0}^{n-1}b^ka^{n-1-k} \end{align}$$ It suffices to show that $$b^n-1 = (b-1)\sum_{k=0}^{n-1}b^k,$$ as follows: $$\begin{align} (b-1)\sum_{k=0}^{n-1}b^k &=b\sum_{k=0}^{n-1}b^k-\sum_{k=0}^{n-1}b^k\\ &=\sum_{k=0}^{n-1}b^{k+1}-\sum_{k=0}^{n-1}b^k\\ &=\sum_{k=1}^{n}b^{k}-\sum_{k=0}^{n-1}b^k\\ &=b^{n}-1\\ \end{align}$$

Aysha A.
  • 660