# Proving $a^n-b^n=(a-b)\sum_{k=1}^{n}a^{n-k} b^{k-1}$ with induction [duplicate]

How can I prove this using induction. I proved for n=1 but now I'm feeling confused while I'm trying to prove for n+1 because of how the summation develops

$$a^n-b^n=(a-b)\sum_{k=1}^{n}a^{n-k} b^{k-1}$$

NOTE: I really need to use induction to prove this, that's why the answer of the other question doesn't fit my question...

• Duplicate of $a^n-b^n=(a-b) (\sum_{i=0}^{n-1}a^ib^{n-1-i})$, which itself was marked as a duplicate of Prove that $(a-b) \mid (a^n-b^n)$. Oct 3, 2015 at 16:55
• see $a^n-b^n$ as a polynomial of $\mathbb R[a]$ where $b$ is a root. Therefore, you can factorise $a^n-b^n$ by $a-b$ what gives $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+b^{n-2}a+b^{n-2}).$$
– Surb
Oct 5, 2015 at 11:16

Write $a^{n+1}-b^{n+1}=(a+b)(a^n-b^n)-ab(a^{n-1}-b^{n-1})$ and then apply the induction hypothesis.
Hint: Write $\;a^{n+1}-b^{n+1}=a(a^n-b^n)+ab^n-b^{n+1}$ and apply the induction hypothesis.
However, the most natural way consists in proving first the formula: $$1-x^n=(1-x)(1+x+\dots+x^{n-1})$$ by a very easy induction (actually it is one of the most illuminating examples of induction when one wants to explain it to beginners).
Then set $x=\dfrac ab$.