# 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.

• No need of using induction here. Just direct proof. – Andrea Mori Oct 17 '17 at 12:59
• – Simply Beautiful Art Oct 17 '17 at 13:00
• @SimplyBeautifulArt I don't see that as a duplicate. – 5xum Oct 17 '17 at 13:01
• @5xum Divide the equation here by $a-b$ on both sides, and apply induction. That is what the dupe is, so I think it answers the question. Unless I am misunderstanding? – Simply Beautiful Art Oct 17 '17 at 13:02
• @SimplyBeautifulArt The linked answer has one variable, this question has two... not a major difference, sure, but hardly one being a duplicate... – 5xum Oct 17 '17 at 13:06

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 $(n+1)$

$(a-b)\cdot\left(\sum\limits_{k=0}^{n+1}a^{n+1-k}\;b^k\right) = (a-b)\cdot\left(\sum\limits_{k=0}^{n}a^{n+1-k}\;b^k + b^{n+1}\right) = (a-b)\cdot\left(a\sum\limits_{k=0}^{n}a^{n-k}\;b^k + b^{n+1}\right) = a\,(a-b)\cdot\left(\sum\limits_{k=0}^{n}a^{n-k}\;b^k\right)+(a-b)\cdot b^{n+1} = a\,(a^{n+1}-b^{n+1})+a\,b^{n+1} - b^{n+2} = a^{n+2}-b^{n+2}$

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: $$1-x^{n+2}=(1-x^{n+1})+(x^{n+1}-x^{n+2})=\dotsm$$

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}

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.

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....

$$(a-b)\sum_{k=0}^{n}a^{n-k}b^{k}= a^{n+1}-b^{n+1}$$

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

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Guy Fsone Oct 17 '17 at 13:44
• I suggester another way to prove the identity. – Suprabha Oct 18 '17 at 6:28
• @GuyFsone is my approach wrong or useless? – Suprabha Oct 21 '17 at 17:04