kostyan5 wrote:

I used mostly brute force, with some factoring, to solve this. I'd be interested in a more elegant solution.

\(\frac{3*(4^5+4^6+4^7+4^8+4^9+4^{10})}{(2^5+2^6+2^7+2^8+2^9+2^{10})} = \frac{3*4^5(1+4^1+4^2+4^3+4^4+4^5)}{2^5*(1+2^1+2^2+2^3+2^4+2^5)} = \frac{3*2^5*2^5(1+4+16+64+256+1024)}{2^5*(1+2+4+8+16+32)} =\)

\(=\frac{3*2^5*1365}{63} = \frac{2^5*1365}{21} = 2^5*65 = 2^5*(64+1) = 2^5*(2^6+1) = 2^{11}+2^5\)

Yes, that is the way I was going at it too, but I realised that there has to be an easier way to do this under 2 minutes.

\(\frac{3*(4^5+4^6+4^7+4^8+4^9+4^{10})}{(2^5+2^6+2^7+2^8+2^9+2^{10})}\)

just consider the top and bottom halves of the geometric progression.

S1 = \((4^5+4^6+4^7+4^8+4^9+4^{10})\)

and S2 = \((2^5+2^6+2^7+2^8+2^9+2^{10})\)

Solving for S1, since the common fraction in the GP is 4, multiply S1 by 4.

Thus, 4S1 = \(4*(4^5+4^6+4^7+4^8+4^9+4^{10}) = (4^6+4^7+4^8+4^9+4^{10}+4^{11})\)

Subtract 4S1 - S1 = 3S1 = \((4^{11} - 4^5)\)

or S1 = \(\frac{(4^{11} - 4^5)}{3}\)

Similarly, for S2, multiply by common factor (2) and subtract:

2S2 - S2 = \((2^{11} - 2^5)\)

Notice that 3S1 is the numerator cancels with the 3 of S1.

Thus we get \(\frac{3 *(4^{11} - 4^5)}{3} * \frac{1}{(2^{11} - 2^5)}\).

Solving and factoring:

\(\frac{(4^{11} - 4^5)}{(2^{11} - 2^5)} = \frac{((2^2)^{11} - (2^2)^5)}{(2^{11} - 2^5)} = \frac{((2^{11})^2 - (2^5)^2)}{(2^{11} - 2^5)}\)

Now you see that the numerator takes the form \(a^2 - b^2 = (a+b) (a-b)\)

Solve and you get answer (C) \(2^{11} + 2^5\)