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

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praveenvino
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3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 [#permalink] New post   03 Feb 2011, 16:16
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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
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B
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Bunuel
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Re: Sum of threes [#permalink] New post   04 Feb 2011, 06:32
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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


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.

Answer: B.
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Re: Sum of threes [#permalink] New post   03 Feb 2011, 16:20
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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



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.
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Re: Sum of threes [#permalink] New post   03 Feb 2011, 20:41
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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


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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Re: Sum of threes [#permalink] New post   03 Feb 2011, 21:18
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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
=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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Re: Sum of threes [#permalink] New post   04 Feb 2011, 09:00
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.

Answer: B.


Good one Bunuel....!!

Thanks fluke and Karishma ...
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Re: Sum of threes [#permalink] New post   20 Feb 2011, 09:39
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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.

Answer: B.
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Re: Sum of threes [#permalink] New post   09 Dec 2013, 09:04
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)
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Re: Sum of threes [#permalink] New post   10 Dec 2013, 03:34
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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:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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] New post   19 Apr 2014, 22:04
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] New post   17 Jan 2019, 07:15
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

~ Answer B.
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Re: 3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 [#permalink] New post   22 May 2020, 05:35
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


sum=[2.3^2…2.3^7] is sum=geoprog
geoprog=a(r^n-1)/r-1,
a=2.3^2, r=2.3^3/2.3^2=3, n=6
2.3^2(3^6-1)/3-1=9(3^6-1),
3+3+3+[gp]=9+9(3^6)=9(1+3^6-1),
sum=3^2(3^6)=3^8

ans (B)
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Re: 3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 [#permalink] New post   22 May 2020, 06:44
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

9 + 2*3(3 +3^2+......+3^6)= 9 + 6*3(3^6 -1)/3-1=9 +9(3^6 -1) = 9(1+3^6-1) = 3^2*3^6 = 3^8
B is the answer.
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Re: 3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 [#permalink] New post   12 Nov 2020, 10:03
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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


3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7
Adding the marked 3s we get 2x3 and the exponent of the 3 is 1 so 2x3^1

Using Geometric progression sum formula.
3+2x3^1+2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7
3+2[3^1+3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7]
3+2[3(\(3^7\)-1]/3-1
3+2[3(\(3^3\)-1)]/2
3+\(3^8\)-3
\(3^8\)

Ans. B
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Re: 3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 [#permalink] New post   09 Feb 2023, 05:37
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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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