How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$.

Question

Please tell if the problem can be solved using telescoping technique or not.

If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, n \in \mathbb{N}.$

I tried as follows, but was unsuccessful to pursue:

$a^n − b^n = a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n=a^n+\sum_{i=1}^{n-1}a^ib^{n-i}-\sum_{i=1}^{n-1}a^ib^{n-i}-b^n$

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Edit : based on the selected answer's comment.

Writing a few terms of the series, $\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i)$ get:

For $n =5$, get the terms as:
$i=1, \,\, a^{5+1-1}b^{1-1}-a^{5-1}b^1 = a^5-a^4b.$
$i=2, \,\, a^{5-1}b^{2-1}-a^{5-2}b^2 = a^4b-a^3b^2.$
$i=3, \,\, a^{5-2}b^{3-1}-a^{5-3}b^3 = a^3b^2-a^2b^3.$
$i=4, \,\, a^{5-3}b^{4-1}-a^{5-4}b^4 = a^2b^3-a^1b^4.$
$i=5, \,\, a^{5-4}b^{3-1}-a^{5-3}b^5 = a^1b^4-b^5.$

Adding all the terms, get:

$a^5-a^4b+ a^4b-a^3b^2+a^3b^2-a^2b^3+a^2b^3-a^1b^4+a^1b^4-b^5 = a^5 - b^5$

Answer 1

\begin{align} (a-b)\sum_{i=1}^n (a^{n-i}b^{i-1}) &=\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i )\\ &=a^n+\sum_{i=2}^n a^{n+1-i}b^{i-1}-\sum_{i=1}^{n-1}a^{n-i}b^i - b^n \\ &= a^n+\sum_{i=1}^{n-1}a^{n-i}b^i-\sum_{i=1}^{n-1}a^{n-i}b^i-b^n\\ &=a^n-b^n \end{align}

You might like to read the working backward to be similar to what you attempted.

$b<a$, then

$$b^{i-1}\le a^{i-1}$$

$$a^{n-i}b^{i-1}\le a^{n-1}$$

$$\sum_{i=1}^na^{n-i}b^{i-1}\le \sum_{i=1}^na^{n-1}=na^{n-1}$$

Answer 2

For the equalty,

By telescoping: We have,

$$(a-b)\sum_{i=1}^na^{n-i}b^{i-1}=a\sum_{i=1}^na^{n-i}b^{i-1}-b\sum_{i=1}^na^{n-i}b^{i-1},$$ which can be rewritten $$\sum_{i=1}^na^{n+1-i}b^{i-1}-\sum_{i=1}^na^{n-i}b^i=a^n+\sum_{i=2}^na^{n+1-i}b^{i-1}-b^n-\sum_{i=1}^{n-1}a^{n-i}b^i,$$ which on simplifying gives $$a^n-b^n+\sum_{i=1}^{n-1}a^{n-i}b^i-\sum_{i=1}^{n-1}a^{n-i}b^i=a^n-b^n.$$ By induction: You wish to prove $$a^n-b^n=(a-b)\sum_{i=1}^n a^{n-i}b^{i-1}.$$ Let's try induction. For the $n=1$ case we have

$$(a-b)a^0b^0=a^1-b^1,$$ so the base case holds.

Now suppose the general case is true and consider the $n+1$ case. We have

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

which can be rewritten $$a(a^n-b^n)+(a-b)b^n=a^{n+1}-ab^n+ab^n-b^{n+1}=a^{n+1}-b^{n+1},$$ so in fact the general case holds.

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For the inequality, you can use induction too:

$$a^n-b^n\leq (a-b)na^{n-1}.\tag{*}$$ The base case clearly holds since $a^1-b^1\leq (a-b)a^0$. Now suppose (*) holds and consider the $n+1$ case,

$$a^{n+1}-b^{n+1}=a^na-b^nb=(a^n-b^n)(a+b)-a^nb+b^na\tag{1}$$

But since $a>b$, then

$$(1)\leq (a^n-b^n)(a+b)-a^nb+a^nb=(a^n-b^n)(a+b)-b(a^n-b^n),$$ which we can write $$(a^n-b^n)a\leq a(a-b)na^{n-1}=(a-b)na^n\leq(a-b)(n+1)a^n,$$ as required.

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Answer 3

Geometric progression \begin{align} (a-b)\sum_{i=1}^{n}a^{n-i}b^{i-1}&=(a-b)a^{n-1}\left(\frac{1-\left(\frac{b}{a}\right)^n}{1-\frac{b}{a}}\right)\\ &=a^n-b^n. \end{align} Can you complete the answer now?

Answer 4

Hint: Use that $$\sum_{i=1}^na^{n-i}b^{i-1}=\frac{a^n-b^n}{a-b}$$ if $$a\ne b$$