devmillenium2k wrote:

If \(R =\frac{30^{65} - 29^{65}}{30^{64} + 29^{64}}\) then

(A) \(0<R\leq{0.1}\)

(B) \(0.1<R\leq{0.5}\)

(C) \(0.5<R\leq{1.0}\)

(D) R<1.0

(E) R>1.0

\(a^n – b^n = (a – b)(a^{n-1} - a^{n-2}*b + a^{n-3}*b^2 …… + b^{n-1})\)

Plugging in our values:

\(1*(30^{64} - 30^{63}*29 + 30^{62}*29^2 - ….. + 29^{64}) = 30^{64} + 29^{64} + k, k > 1.\)

Hence \(30^{65} – 29^{65} = 30^{64} + 29^{64} + k > 30^{64} + 29^{64}\)

In general: \(a^n - b^n = a^{n-1} + b^{n-1} +k > a^{n-1} + b^{n-1}\), because \(k\) is always \(> 1\)

And our fraction is \(> 1\).

If power of polynomial in numerator is greater than power of polynomial in denominator the value of fraction will always be >1. We can prove this with application of simple calculus. Just by taking consecutive derivatives of numerator and denominator.

\(\frac{65*64*63*62* ….. *x}{64*63*62* …*1}\)

Which is always \(> 1.\)