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Intern  Joined: 06 Nov 2010
Posts: 19
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6  [#permalink]

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9 00:00

Difficulty:   65% (hard)

Question Stats: 65% (02:24) correct 35% (02:58) wrong based on 372 sessions

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3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: Sum of threes  [#permalink]

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7
praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7
(B) 3^8
(C) 3^14
(D) 3^28
(E) 3^30

Here is even faster and smarter approach:

We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*3^7 we would have: sum=9*(2*3^7)=2*3^9=~3^10, so the actual sum is less than 3^10 and more than 3^7 (option A) as the last term is already more than that. So the answer is clearly B.

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Intern  Joined: 06 Nov 2010
Posts: 19
Re: Sum of threes  [#permalink]

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1
praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30

OE:

The key to this problem is realizing that 3 + 3 + 3 = 9 = 3^2. Let us rewrite the prompt accordingly: 3^2 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7.

Now, since this equation contains exponents, we should try to manipulate the terms so that we can use the exponent rules we know. We can rewrite the first three terms as 3^2 + 3^2 + 3^2. This is equivalent to 3(3^2) = 3^1(3^2) = 3^3.

The equation can thus be simplified: 3^3 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7.

Since 3 × 3^3 = 3^4, we can simplify further: 3^4 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7.

We repeat this procedure until we get: 3 × 3^7 = 38.

Answer choice B is correct.
General Discussion
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: Sum of threes  [#permalink]

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2
4
praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30

Faster approach using number properties:

$$2^0 + 2^1 + 2^2 + .... 2^{n - 1} = 2^n - 1$$
$$2( 3^0 + 3^1 + 3^2 + .... + 3^{n - 1}) = 3^n - 1$$
$$3 (4^0 + 4^1 + 4^2 + ....+ 4^{n - 1}) = 4^n - 1$$
and so on...

Given Question:
$$3 + 3 + 3 + 2.3^2 + 2.3^3 + 2.3^4 + 2.3^5 + 2.3^6 + 2.3^7$$
$$3 + 2.3 + 2.3^2 + 2.3^3 + .... + 2.3^7$$
$$3 + 2(3^1 + 3^2 + .... + 3^7)$$
$$3 + 2.[(3^8 - 1)/2 - 1] = 3^8$$
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Retired Moderator Joined: 20 Dec 2010
Posts: 1579
Re: Sum of threes  [#permalink]

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1
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7
=3^2+ 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7
=3^2+2*3(3+3^2+3^3+3^4+3^5+3^6)
###(3+3^2+3^3+3^4+3^5+3^6) = ((3^6)-1)/(3-1) = 3*(((3^6)-1)/2)###
=3^2+2*3*3*((3^6)-1)/2 =3^2+3^8-3^2=3^8

Ans: "B"

The idea is to change the above series as a geometric series or part Geometric series;

3,3^2,3^3,3^4,3^5,3^6 are in geometric progression(G.P.).

a=first element of the series=3
r=ratio between two neighboring terms=3^2/3=3
n=number of elements=6

Sum of the G.P. = $$a(r^n-1)/(r-1)$$ if r>1.

Since r=2>1

Sum of the G.P. of the above series= $$3(3^6-1)/(3-1)$$
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Intern  Joined: 06 Nov 2010
Posts: 19
Re: Sum of threes  [#permalink]

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Bunuel wrote:
praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7
(B) 3^8
(C) 3^14
(D) 3^28
(E) 3^30

Here is even faster and smarter approach:

We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*3^7 we would have: sum=9*(2*3^7)=2*3^9=~3^10, so the actual sum is less than 3^10 and more than 3^7 (option A) as the last term is already more than that. So the answer is clearly B.

Good one Bunuel....!!

Thanks fluke and Karishma ...
Manager  Joined: 17 Feb 2011
Posts: 143
Concentration: Real Estate, Finance
Schools: MIT (Sloan) - Class of 2014
GMAT 1: 760 Q50 V44 Re: Sum of threes  [#permalink]

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1
Very good approach.

Bunuel wrote:
praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7
(B) 3^8
(C) 3^14
(D) 3^28
(E) 3^30

Here is even faster and smarter approach:

We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*3^7 we would have: sum=9*(2*3^7)=2*3^9=~3^10, so the actual sum is less than 3^10 and more than 3^7 (option A) as the last term is already more than that. So the answer is clearly B.

Intern  Joined: 18 Sep 2013
Posts: 22
Re: Sum of threes  [#permalink]

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VeritasPrepKarishma wrote:
praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30

Faster approach using number properties:

$$2^0 + 2^1 + 2^2 + .... 2^{n - 1} = 2^n - 1$$
$$2( 3^0 + 3^1 + 3^2 + .... + 3^{n - 1}) = 3^n - 1$$
$$3 (4^0 + 4^1 + 4^2 + ....+ 4^{n - 1}) = 4^n - 1$$
and so on...

Given Question:
$$3 + 3 + 3 + 2.3^2 + 2.3^3 + 2.3^4 + 2.3^5 + 2.3^6 + 2.3^7$$
$$3 + 2.3 + 2.3^2 + 2.3^3 + .... + 2.3^7$$
$$3 + 2(3^1 + 3^2 + .... + 3^7)$$
$$3 + 2.[(3^8 - 1)/2 - 1] = 3^8$$

Hi Karishma,
Could you please explain the last step. How did you get 3^8 and divided by (2-1)
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: Sum of threes  [#permalink]

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1
Sam1 wrote:
VeritasPrepKarishma wrote:
praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30

Faster approach using number properties:

$$2^0 + 2^1 + 2^2 + .... 2^{n - 1} = 2^n - 1$$
$$2( 3^0 + 3^1 + 3^2 + .... + 3^{n - 1}) = 3^n - 1$$
$$3 (4^0 + 4^1 + 4^2 + ....+ 4^{n - 1}) = 4^n - 1$$
and so on...

Given Question:
$$3 + 3 + 3 + 2.3^2 + 2.3^3 + 2.3^4 + 2.3^5 + 2.3^6 + 2.3^7$$
$$3 + 2.3 + 2.3^2 + 2.3^3 + .... + 2.3^7$$
$$3 + 2(3^1 + 3^2 + .... + 3^7)$$
$$3 + 2.[(3^8 - 1)/2 - 1] = 3^8$$

Hi Karishma,
Could you please explain the last step. How did you get 3^8 and divided by (2-1)

There are two ways to get that:
1. The number property discussed above: $$2( 3^0 + 3^1 + 3^2 + .... + 3^{n - 1}) = 3^n - 1$$
If n = 8,
$$2( 3^0 + 3^1 + 3^2 + .... + 3^7) = 3^8 - 1$$
$$3^0 + 3^1 + 3^2 + .... + 3^7 = (3^8 - 1)/2$$
Since 3^0 = 1,
$$1 + 3^1 + 3^2 + .... + 3^7 = (3^8 - 1)/2$$
$$3^1 + 3^2 + .... + 3^7 = (3^8 - 1)/2 - 1$$
That's how we substituted $$(3^8 - 1)/2 - 1$$ for $$3^1 + 3^2 + .... + 3^7$$

Another method is using the GP formula
$$3^1 + 3^2 + .... + 3^7 = 3(3^7 - 1)/(3 - 1) = (3^8 - 3)/2 = (3^8 - 1 - 2)/2 = (3^8 - 1)/2 - 1$$
Here is a post explaining GPs:
http://www.veritasprep.com/blog/2012/04 ... gressions/
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Re: 3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6  [#permalink]

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The above can be changed to a geometric progression.

3+3+3+2*3^2+2*3^3+.....................+2*3^6

9 + 2( 3^2+...........+3^6)

Sum for a geometric progression given below.

a+ar+ar^2+................+ar^n is

S= a(r^n-1)/(r-1)

Applying this to our series.

9 + 2 { 9 (3^6 - 1)/(3-1)}

=> 3^8
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3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6  [#permalink]

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praveenvino wrote:
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 =

(A) 3^7

(B) 3^8

(C) 3^14

(D) 3^28

(E) 3^30

Somehow i took time while solving this question, but once i realized the pattern, it was not difficult per se

3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7
3^2 + 2 * 3^2 [ 1 + 3 + 9 + 27 + 81 + 243]

Now there is a GP, for which we can calculate the sum Sn = a r ^ (n -1)/ r - 1

there are 6 terms present, S6 = 3^5/2

3^2 + 3^7

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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up. 3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6   [#permalink] 17 Jan 2019, 07:15
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