Proof that: $(a-b)\cdot\Bigg(\sum_{k=0}^{n}a^{n-k}b^{k}\Bigg)=a^{n+1}-b^{n+1}\text{ }\forall n\in\mathbb{N_{0}}$ I'm trying to prove a more general version of the 3rd binomial equation via mathematical induction which will help me complete another proof.
$$(a-b)\cdot\Bigg(\sum_{k=0}^{n}a^{n-k}b^{k}\Bigg)=a^{n+1}-b^{n+1}\text{ }\forall n\in\mathbb{N_{0}}$$
I proved the base case but I'm unable to make progress in the inductive step.
 A: We first show it works for $n = 1$:
$(a-b)\cdot (a+b) = a^2-b^2$
Now, suposing it works for $(n)$, let's show it works for $(n+1)$
$(a-b)\cdot\left(\sum\limits_{k=0}^{n+1}a^{n+1-k}\;b^k\right) = (a-b)\cdot\left(\sum\limits_{k=0}^{n}a^{n+1-k}\;b^k + b^{n+1}\right) = (a-b)\cdot\left(a\sum\limits_{k=0}^{n}a^{n-k}\;b^k + b^{n+1}\right) = a\,(a-b)\cdot\left(\sum\limits_{k=0}^{n}a^{n-k}\;b^k\right)+(a-b)\cdot b^{n+1} = a\,(a^{n+1}-b^{n+1})+a\,b^{n+1} - b^{n+2} = a^{n+2}-b^{n+2}$
A: Hint:
Begin with the non-homogeneous case
$$1-x^{n+1}=(1-x)(1+x+\dots+x^n)\tag{1}$$
then factor out $a^{n+1}$ and set $x=\dfrac ba$.
Inductive step:
You have to  prove 
$$1-x^{n+2}=(1-x)(1+x+\dots+x^{n+1})$$
from the hypothesis that $(1)$ is true for some $n$, which is easy:
$$1-x^{n+2}=(1-x^{n+1})+(x^{n+1}-x^{n+2})=\dotsm$$
A: Let me start for you:
$$\begin{align}(a-b)\cdot\sum_{k=0}^n a^{n-k}b^k&=a\cdot\sum_{k=0}^n a^{n-k}b^k -
 b\cdot \sum_{k=0}^n a^{n-k}b^k\\
&=\sum_{k=0}^na\cdot  a^{n-k}b^k-\sum_{k=0}^n b\cdot a^{n-k}b^k\\
&=\sum_{k=0}^n a^{n-k+1}b^k - \sum_{k=0}^n - a^{n-k}b^{k+1}\\
& = \sum_{k=1}^{n} a^{n-k+1}b^k + a^{n+1} -\left(\sum_{k=0}^{n-1} a^{n-k}b^{k+1} + b^{n+1}\right)\end{align}$$
A: It's a geometric serie that start with $a^n$ and ends with $b^n$ the ratio  is $ \frac b a$.
The evaluation of the serie is therefore easy:
$$\sum_{k=0}^{n}a^{n-k}b^{k}=\frac {\frac {b^{n+1}} a -a^n}{\frac b a -1}=\frac {b^{n+1}- a^{n+1}} {b-a}$$
Then simplify....
$$(a-b)\sum_{k=0}^{n}a^{n-k}b^{k}= a^{n+1}-b^{n+1}$$
A: Hint: 
Use sum of GP with 1st term $a^n$ and common ratio $b/a$ till $b^n$.
$a^n+a^{n-1}×b+a^{n-2}×b^2+... b^n$
Its value is $ a^n(1-(b/a)^{n+1})/(1-b/a)$. From which we can reach the required result. The sum of GP can also bo proved much easily
A: Say $$ I = a^n+a^{n-1}b+a^{n-2}b^2+...+a^2b^{n-2}+ab^{n-1}+b^{n}$$
 Then $$aI = a^{n+1}+
\underbrace{a^{n}b+a^{n-1}b^2+...+a^3b^{n-2}+a^2b^{n-1}+ab^{n}}$$
and $$bI = 
\underbrace{a^nb+a^{n-1}b^2+a^{n-2}b^3+...+a^2b^{n-1}+ab^{n}}+b^{n+1}$$
so we have $$aI-bI = a^{n+1}-b^{n+1}$$
and we are done.
