Which is larger, where n is a natural number:
99^{n} + 100^{n} or 101^{n} ?
(In reply to
Near Analytic Solution by K Sengupta)
Nice work, KS. The proof you asked for:
Evaluating (101^n  99^n)/100^n, we get
2* (n/100 + n(n1)(n2)/(3!*100^3) + ...
You already shown that this expression (by your means) exceeds 1 for n = 50 and for n = 49.
To show that the expression is less than 1 for n = 48:
A = 2*{48/100 + (48*47*46)/(3!*100^3) + (48*47*46*45*44)/(5!*100^5) + ...} is less than
2*[48/100 + (48^3)/(1*2*3)*100^3) + (48^5)/(1*2*3)(2*3)*100^5) + (48^7)/(1*2*3)(2*3)(2*3)*100^7 + ...]
where [x] is the greatest integer less than x.
= 2*[48/100 + 1/6*(48/100)^3 + 1/(6^2)*(48/100)^5 +...]
< 2* {(48/100) / (1  1/6*(48/100)^2}
= 9600/9616 < 1.
This problem was posed in a USSR Math Olympiad, in the lates 40īs, when there wasnīt electronic spreadsheet available, yet.
Edited on June 30, 2008, 1:41 pm

Posted by pcbouhid
on 20080630 13:30:07 