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## Homework Statement

let

*n*[tex]\in[/tex]

**N**To prove the following inequality

*n*a[tex]^{n-1}[/tex](b-a) < b[tex]^{n}[/tex] - a[tex]^{n}[/tex] <

*n*b[tex]^{n-1}[/tex](b-a)

0<a<b

## Homework Equations

## The Attempt at a Solution

Knowing that b^n - a^n = (b-a)(b^(n-1) + ab^(n-2) + ... + ba^(n-2) + a^(n-1) we can divide out (b-a) because b-a # 0. so we have

n(a^(n-1)) < (b^(n-1) + ab^(n-2) + ... + ba^(n-2) + a^(n-1) < n(a^(n-1))

and in the middle of the inequality we see that

na^(n-1) < a^(n-1)< ab^(n-2) + ... + ba^(n-2) < a^(n-1) <nb^(n-1)

and in the middle there are a^j (b^(n-1-j)) terms as j --> n-2

So there are (n-1)+1 terms in the middle or n terms. So if a<b this inequality holds....

I think i pulled some random stuff out of thin air, but its a try.. THanks