<s>[INST] Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) = \frac{b^{n + 1} - a^{n + 1}}{(b-a)}$
What can I do from here?
 [/INST] Hint
We have that 
$$\dfrac{b^{n+1}-a^{n+1}}{b-a}=b^n+b^{n-1}a+b^{n-2}a^2+\cdots +a^2b^{n-2}+ba^{n-1}+a^n$$
Is not $a^n< b^ia^{n-i}< b^n, \forall i\in \{0,\dots, n\}?$ Are not there $n+1$ summands?
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