Show that if 0<=a<b, then (b^(n+1) - a^(n+1))/(b - a) < (n+1)*b^n...?

Answer 1

The quotient (b^(n+1) - a^(n+1))/(b - a) is equal to
b^n + b^(n-1)*a + b^(n-2)*a^2 + ... + b*a^(n-1) + a^n. There are n + 1 terms on the right, each of the form b^(n-t)*a^t. Since 0 <= a < b, each b^(n-t)*a^t < b^(n-t)*b^t = b^n, and there are n + 1 such terms, which completes the proof.

Answer 2

This difficulty has different blind alleys, yet one thank you to teach this is to contemplate a rational fraction of the kind: a/(an-b) the place n > a > b are integers. Subtracting a million/n from it, we've: a/(an-b) - a million/n = (an - an + b)/(n(an-b)) = b/(an² - bn) the place we replace an² - bn with bn' - c, the place b > c. for this reason, we've a the rest fraction the place b < a, and this technique could be repeated till we are left with a fragment of the kind a million/n'', and the algorithmic technique ends with a finite sum of fractions. it may be observed that for any given rational fraction 0 < x < a million, there's a multiplicity of techniques it extremely is expressed interior the kind a million/a + a million/b + a million/c + ..., the place a, b, c... are all different integers. This difficulty jogs my memory of Turing's Halting difficulty, in that what we are being asked to do is to teach that if we subtract fractions of the kind a million/n successively from the unique rational fraction, the approach will halt finally. this is style of exciting to objective doing it utilising random fractions, finally it halt with now and lower back some notably some distance out fractions. Addendum: If for any reason in the process the algorithmic technique we arise with 2 fractions that are alike, we can use amir11elad's approach of adjusting between the fractions with 2 others, and if that motives yet another tournament, we can repeat, till after a finite form of step there will be no extra suits and all the fractions could be different. Addendum 2: right here is the evidence lower back, greater concisely: a million) Subtract fractions of the kind a million/n from the rational selection 0 < x < a million successively (with different n) till we are left with a fragment of the kind a/(an-b), the place n > a > b are integers. 2) Subtract a million/n from this fraction, so as that we are left with a fragment of the kind b/(bn-c), the place n > b > c are integers. 3) proceed till we are left with the fraction of the kind a million/n. in case you have a chain of integers a > b > c > ...., finally, you will finally end up with a million. it relatively is why the sequence will consistently be finite. Addendum 3: pay attention ye, Alexey V