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Math Expert V
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In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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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.

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In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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

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Re: In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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I followed a very manual approach:

The formula given is simple so populated a few numbers:

G1 =1
G2= 1
G3= 1+ 2^(3-3) = 1+1 = 2
G4= 2+ 2^ (4-3) = 2+2 = 4
G5= 4+2^(5-3)=2+4 = 8
G6=8+ 2^(6-3)=16

The trends shows that numbers will be in powers of 2 - so G7 will be 16*2 = 32, G8 = 32*2 = 64

Next I took an approximation of sum of the series and here is what the trend suggested:

1. G1+G2+G3= G4
2. G1+ G2+G3+G4+G5+G6+G7 =G8

So basically all numbers add up to to give the next in series.

This implies - Sum of G1 upto G16 will be 16384 and the sum of series upto G15 is 8192.

hence, the solution is option A - i.e. 2

Please suggest if this approach is fine.
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In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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aimtoteach wrote:
I followed a very manual approach:

The formula given is simple so populated a few numbers:

G1 =1
G2= 1
G3= 1+ 2^(3-3) = 1+1 = 2
G4= 2+ 2^ (4-3) = 2+2 = 4
G5= 4+2^(5-3)=2+4 = 8
G6=8+ 2^(6-3)=16

The trends shows that numbers will be in powers of 2 - so G7 will be 16*2 = 32, G8 = 32*2 = 64

Next I took an approximation of sum of the series and here is what the trend suggested:

1. G1+G2+G3= G4
2. G1+ G2+G3+G4+G5+G6+G7 =G8

So basically all numbers add up to to give the next in series.

This implies - Sum of G1 upto G16 will be 16384 and the sum of series upto G15 is 8192.

hence, the solution is option A - i.e. 2

Please suggest if this approach is fine.

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

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In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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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.

We want $$\frac{SUM(g_1...g_{16})}{SUM(g_1..g_{15})}$$
$$g_1=g_2=1$$
$$g_3=g_2+2^{3-3}=1+2^0=1+1=2$$
$$g_4=2+2=4$$
$$g_5=4+4=8$$
$$g_n=2^{n-2}$$ for $$n\geq{2}$$

$$ψ(g_{1})=1$$
$$ψ(g_{2})=1+1=2$$
$$ψ(g_{3})=2+2=4$$
$$ψ(g_{4})=4+4=8$$
$$ψ(g_{n})=2^{n-1}$$ for $$n\geq{1}$$

Therefore, $$\frac{ψ(g_{16})}{ψ(g_{15})}=\frac{2^{15}}{2^{14}}=2$$.
Of the answer choices. $$A=2$$ so the answer is A.
Math Expert V
Joined: 02 Sep 2009
Posts: 58454
Re: In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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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.

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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GMAT 1: 760 Q50 V42 Re: In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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Bunuel

How much time should one allot to these kinds of questions, I don't think it solvable in 2 minutes.

what do you suggest?
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Math Expert V
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Re: In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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DIII wrote:
Bunuel

How much time should one allot to these kinds of questions, I don't think it solvable in 2 minutes.

what do you suggest?

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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Re: In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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DIII wrote:
Bunuel

How much time should one allot to these kinds of questions, I don't think it solvable in 2 minutes.

what do you suggest?

The very important first point has already been mentioned by bunuel which is Average time 2 mins per question does NOT mean you have to finish every question in 2 mins. There are always some question that can be solved in less than a min then save time there and use that time in harder questions which deserve 3 mins or even more

Second important point that such lengthy questions are highly unlikely to appear to GMAT. If you can't see and short solution then understand that such questions are good for practice but unlikely to be seen in GMAT
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GMAT 1: 760 Q50 V42 Re: In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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Thank you GMATinsight Bunuel
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Re: In the sequence gn defined for all positive integer values of n, g1 =  [#permalink]

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