Which is larger, where n is a natural number:
99n + 100n or 101n ?
At the outset, in conformity with this definition in this location, we take that a natural number is a nonnegative integer.
Now, substituting A(n) = 1.01^n - 0.99^n, we observe that:
101^n is > or < (99^n + 100^n), accordingly as:
A(n) is > or < 1
Also, we must have:
A(n) - A(n-1) = 0.01(1.01^(n-1) + 0.99^(n-1)), where n is a nonnegative integer.
Thus, A(n) is increasing in n.
Therefore, if A(n) > 1, then A(n+t) > 1, for nonnegative integers t, and:
If A(n) < 1, then A(n-t) < 1, for nonnegative integers t.
Now, expanding by Binomial Theorem, we have:
A(n) = 1.01^n - 0.99^n
= (1 + 1/100)^n - (1 - 1/100)^n
= n/50 + n(n-1)(n-2)/(3*100^3) + ......
> 1, whenever n >= 50
= 49/50 + (36848)/(100^3) + .....
> 0.98 + 0.036848 = 1.016848 > 1
Accordingly, A(n) > 1, for n >= 49
Also, it can be verified by calculator that:
A(48) = 1.01^48 - 0.99^48 < 1
Therefore, from (#), it follows that:
A(n) < 1, whenever n <= 48
101^n > 99^n + 100^n, whenever n>=49, and:
101^n < 99^n + 100^n, whenever 0<= n<= 48
It may be observed that this methodology is almost analytic, and all the calculations (except A(48)) were done by hand. Howerver, calculator aid was required at some stage, and accordingly, this solution cannot be termed as fully analytic, and an analytic proof that A(48) < 1 is required to complete the solution.
Edited on June 30, 2008, 4:07 pm