Bunuel wrote:

In the sequence \(g_n\) defined for all positive integer values of n, \(g_1 = g_2 = 1\) and, for n ≥ 3, \(g_n = g_{n–1} + 2^{n-3}\). If the function \(ψ(g_i)\) equals the sum of the terms \(g_1\), \(g_2\), …, \(g_i\) , what is \(\frac{ψ(g_{16})}{ψ(g_{15})}\)

(A) \(g_3\)

(B) \(g_8\)

(C) \(ψ(g_8)\)

(D) \(ψ(g_{16}) - ψ(g_{15})\)

(E) \(\frac{g_{16}}{2}\)

Kudos for a correct solution.

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 2Question : \(\frac{ψ(g_{16})}{ψ(g_{15})}\)\(ψ(g_{16})\) = 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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