If R =(30^65 - 29^65)/(30^64 + 29^64) then

This question is based on your understanding of number properties. You don't have to do any calculations actually.
At higher powers, a difference of even 1 in the base creates creates a huge difference in the number. For example, \(2^{10} = 1024\) but \(3^{10} = 59049\).

So \(30^{65}\) will not be very different from \(30^{65}\)-\(29^{65}\).

Similarly, \(30^{64}\) will not be very different from \(30^{64}\)+\(29^{64}\)

So the question boils down to approx \(\frac{30^{65}}{30^{64}}\). This will be something less than 30 but certainly greater than 1.

Answer (E)
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\(R = \frac{30^{65} - 29^{65}}{30^{64} + 29^{64}} \approx{\frac{30^{65}}{30^{64}}} \approx{30}\)

Answer = E

\(30^{65} or 30^{64}\) is a very big number; any addition/subtraction of "equivalent value" would hardly make a big difference

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

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 \(> 1.\)

You don't need to solve using the exact numbers in the problem-

You can use for example: (3^3 - 2^3)/(3^2 + 2^2) = 19/13

Hence, R>1 --> E is correct

R= 30^{65} - 29^{65}/30^{64} + 29^{64}
=>[ 30^{64} + 29^{64} ]R= 30^{65} - 29^{65}
=> 30^{64} [30-R] = 29^{64} [29+R]
=> (30/29)^{64) = (29+R)/(30-R)
Now, LHS > 1 => 29+R> 30-R => R > 0.5
Therefore, R= ( 0.5, ∞ )

Ans. only E satisfy this condt.

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