Given : \(g_1 = g_2 = 1\) and, for n ≥ 3, \(g_n = g_{n–1} + 2^{n-3}\)

i.e. \(g_3 = g_2 + 2^{3-3} = 1 + 1 = 2 = 2^1\)
i.e. \(g_4 = g_3 + 2^{4-3} = 2 + 2 = 4 = 2^2\)
i.e. \(g_5 = g_4 + 2^{5-3} = 4 + 4 = 8 = 2^3\)
i.e. \(g_6 = g_5 + 2^{6-3} = 8 + 8 = 16 = 2^4\)
and so on...
i.e. \(g_{15} = g_{14} + 2^{15-3} = 2^{13}\)
i.e. \(g_{16} = g_{15} + 2^{16-3} = 2^{14}\)

i.e. Each next term is next higher power of 2

Question : \(\frac{ψ(g_{16})}{ψ(g_{15})}\)

\(ψ(g_{16})\) = sum of the terms \(g_1\), \(g_2\), …, \(g_16\) = \(1+1+2^1+2^2+---+2^14\) = \(2+\frac{2*(2)^{14-1}}{(2-1)}\)

i.e. \(ψ(g_{16})\) = \(2+2*(2)^{13}\) = \(2+(2)^{14}\) = \(2*(1+2^{13})\)

Similarly,
\(ψ(g_{15})\) = sum of the terms \(g_1\), \(g_2\), …, \(g_15\) = \(1+1+2^1+2^2+---+2^13\) = \(2+\frac{2*(2)^{13-1}}{(2-1)}\)

i.e. \(ψ(g_{15})\) = \(2+2*(2)^{12}\) = \(2+(2)^{13}\) = \(2*(1+2^{12})\)

Now,
\(\frac{ψ(g_{16})}{ψ(g_{15})}\) = \(\frac{2*(1+2^{13})}{2*(1+2^{12})}\)

i.e. \(\frac{ψ(g_{16})}{ψ(g_{15})}\) = \(\frac{(1+2^{13})}{(1+2^{12})}\)

But \(2*(1+2^{12})\) is very closely approximately equal to \(2^{12}\)
and \(2*(1+2^{13})\) is very closely approximately equal to \(2^{13}\)

Therefore, i.e. \(\frac{ψ(g_{16})}{ψ(g_{15})}\) = \(\frac{(1+2^{13})}{(1+2^{12})}\) = \(\frac{(2^{13})}{(2^{12})}\) = \(2\)

Answer: Option
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This is absolutely correct however the Last step could be made easier

Given : \(g_1 = g_2 = 1\) and, for n ≥ 3, \(g_n = g_{n–1} + 2^{n-3}\)

i.e. \(g_3 = g_2 + 2^{3-3} = 1 + 1 = 2 = g_1+g_2\)
i.e. \(g_4 = g_3 + 2^{4-3} = 2 + 2 = 4 = g_1+g_2+g_3\)
i.e. \(g_5 = g_4 + 2^{5-3} = 4 + 4 = 8 = g_1+g_2+g_3+g_4\)
i.e. \(g_6 = g_5 + 2^{6-3} = 8 + 8 = 16 = g_1+g_2+g_3+g_4+g_5\)
and so on...
i.e. \(g_{15} = g_{14} + 2^{15-3} = g_1+g_2+g_3+g_4----+g_14\)
i.e. \(g_{16} = g_{15} + 2^{16-3} =g_1+g_2+g_3+g_4----g_{14}+g_{15} = 2*g_{15}\)

Question : \(\frac{ψ(g_{16})}{ψ(g_{15})}\)

\(ψ(g_{16})\) = sum of the terms \(g_1\), \(g_2\), …, \(g_{15}\), \(g_{16}\) = \(g_{16}+g_{16}\) = \(2*g_{16}\)

i.e. \(ψ(g_{16})\) = \(2*g_{16}\)

\(ψ(g_{15})\) = sum of the terms \(g_1\), \(g_2\), …, \(g_15\)

i.e. \(ψ(g_{15})\) = \(g_{16}\)

i.e. \(\frac{ψ(g_{16})}{ψ(g_{15})}\) = \(2*g_{16}/g_{16}\) = \(2\)

Answer: Option
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MANHATTAN GMAT OFFICIAL SOLUTION:

We begin by listing some values of gn, in order to get a sense for how gn progresses:
\(g_1 = 1\)

\(g_2 = 1\)

\(g_3 = g_2 + 2^0 = 1 + 1 = 2 = 2^1\)

\(g_4 = g_3 + 2^1 = 2 + 2 = 4 = 2^2\)

\(g_5 = g_4 + 2^2 = 4 + 4 = 8 = 2^3\)

\(g_6 = g_5 + 2^3 = 8 + 8 = 16 = 2^4\)

We can see that for n ≥ 3, \(g_n = 2^{n–2}\).

Let us now look for a pattern in the sums defined as (gn):
\(ψ(g_3) = g_1 + g_2 + g_3 + = 1 + 1 + 2 = 4 = 2^2\)

\(ψ(g_4) = (g_1 + g_2 + g_3) + g_4 = ψ(g_3) + g_4 = 4 + 4 = 8 = 2^3\)

\(ψ(g_5) = (g_1 + g_2 + g_3 + g_4) + g_5 = ψ(g_4)+ g_5 = 8 + 8 = 16 = 2^4\)

Each value is double the previous value: \(ψ(g_n) = 2 * ψ(g_{n-1})\). This means that:
\(\frac{ψ(g_{16})}{ψ(g_{15})}=\frac{2*ψ(g_{15})}{ψ(g_{15})}=2\)

Now all we need to do is scan the answer choices to find an expression that equals 2. We have already discovered that g3 = 2, so we can select g3 as the answer.

The correct answer is A.
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Ideally you should solve easier questions in less than 2 minutes which will allow you to spend a little bit more on those which are harder. The average time of correct answer on this questions is 3:45 minutes.
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Thank you GMATinsight Bunuel
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