- #1
kbrono
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Homework Statement
let n[tex]\in[/tex]N To prove the following inequality
na[tex]^{n-1}[/tex](b-a) < b[tex]^{n}[/tex] - a[tex]^{n}[/tex] < nb[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