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# 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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Updated on: 18 Jun 2017, 06:44
2
34
00:00

Difficulty:

25% (medium)

Question Stats:

76% (01:42) correct 24% (01:58) wrong based on 990 sessions

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$$2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?$$

A. $$2^9$$

B. $$2^{10}$$

C. $$2^{16}$$

D. $$2^{35}$$

E. $$2^{37}$$

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Originally posted by Baten80 on 13 Aug 2010, 10:15.
Last edited by Bunuel on 18 Jun 2017, 06:44, edited 5 times in total.
Edited the question and added the OA
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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14 Aug 2010, 10:18
14
10
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

A. 2^9
B. 2^10
C. 2^16
D. 2^35
E. 2^37

$$2+2=2*2=2^2$$ (the sum of first and second terms);
$$2^2+2^2=2^3$$ (the sum of previous 2 terms and the third term);
$$2^3+2^3=2^4$$ (the sum of previous 3 terms and the fourth term);
...
The same will continue and finally we'll get $$2^8+2^8=2*2^8=2^9$$, the sum of previous 8 terms and the 9th term.

OR: we can identify that the terms after the first one represent geometric progression.

Sum of the terms of geometric progression is given by: $$Sum=\frac{a*(r^{n}-1)}{r-1}$$, where $$a$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$>1$$.

In our original question we have 2 plus G.P. with 8 terms, so:

$$2+(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)=2+\frac{2*(2^{8}-1)}{2-1}=2^9$$.

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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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01 Apr 2012, 14:24
9
1
there are nine numbers, so group them up in 3.

$$[2+2+2^2]+[2^3+2^4+2^5]+[2^6+2^7+2^8]$$

$$[2^3]+2^3[1+2+4]+2^6[1+2+4]=2^3+2^3*7+2^3*8*7$$

$$2^3*64=2^3*2^6=2^9$$
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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01 Apr 2012, 19:17
2
amitdgr wrote:
$$2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8$$ = ?

* 2^9
* 2^10
* 2^16
* 2^35
* 2^37

See for pattern:
2 + 2 =2^2
2^2+2^2=2^3
2^3 + 2^3=2^4
........
2^8 + 2^8=2^9

wouldnot take more than 30 seconds.

hope this helps..!!
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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02 Apr 2012, 00:46
1
cangetgmat wrote:
I was stuck performing addition of GP.
+1 to you

You can do it using sum of GP as well.

$$2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 1 + (1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8)$$
The terms in brackets make a GP with first term = 1 and common ratio = 2

GP = $$1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8$$
Sum of 9 terms of GP $$= 1(1 - 2^9)/(1-2) = 2^9 - 1$$

Sum of the required series = 1 + sum of GP = $$1 + 2^9 - 1 = 2^9$$

To check out a discussion on both the methods, check out my blog http://www.veritasprep.com/blog/categor ... er-wisdom/
today evening. My this week's post discusses a question very similar to this one.
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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04 Apr 2012, 01:03
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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10 Dec 2014, 00:27
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 =

Just observe the series

The addition of first 2 terms = 3rd term = 2^2

Addition of first 3 terms = 4th term = 2^3

Addition of first 4 terms = 5th term and so on....................

Addition of first 8 terms = 9th term = 2^8

$$2^8 + 2^8 = 2^9$$
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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06 Jul 2015, 21:25
Bunuel wrote:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

A. 2^9
B. 2^10
C. 2^16
D. 2^35
E. 2^37

$$2+2=2*2=2^2$$ (the sum of first and second terms);
$$2^2+2^2=2^3$$ (the sum of previous 2 terms and the third term);
$$2^3+2^3=2^4$$ (the sum of previous 3 terms and the fourth term);
...
The same will continue and finally we'll get $$2^8+2^8=2*2^8=2^9$$, the sum of previous 8 terms and the 9th term.

OR: we can identify that the terms after the first one represent geometric progression.

Sum of the terms of geometric progression is given by: $$Sum=\frac{a*(r^{n}-1)}{r-1}$$, where $$a$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$>1$$.

In our original question we have 2 plus G.P. with 8 terms, so:

$$2+(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)=2+\frac{2*(2^{8}-1)}{2-1}=2^9$$.

Isn't is safe to use the formula of GP rather than using the other approach?
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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15 Jul 2015, 09:02
Hey,

I didn't think of it as a progression, but I started by taking out 2 as a common term.
Taking out the first common term of 2 you end up with 2(1+1+2+2^2+...+2^8).

In the end you end up with 2*2*2*2*2*2*2*2*2 = 2^9.

You can either finish the whole process, taking out 2 as a common term in each step, or you will end up noticing that in every step you always end up with 9 numbers (terms of 2), inside or outside the parentheses (1+1 = 2, so it counts as one number). After taking out another term of two, one more number is added in those outside the parenthesis and one number is removed from the ones inside the parentheses. You have 9 terms, so in the end there will be 9 two's outside the parentheses: 2^9.
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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17 Aug 2016, 23:58
Bunuel

how do we apply 30 sec approach in this question?

I tried this way... We have the sum of 9 terms. Now, if all terms were equal to the largest term 2^ 8 ,we would have: 9∗(2^8)
--- 9 can be written as 2^3 appprox ... so largest term becomes 2^3 ∗ 2^8 = 2^11
So answer has to be less than 2^11

we have two answer options less than 2^11...how to get into right answer?
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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18 Aug 2016, 07:46
kapru wrote:
Bunuel

how do we apply 30 sec approach in this question?

I tried this way... We have the sum of 9 terms. Now, if all terms were equal to the largest term 2^ 8 ,we would have: 9∗(2^8)
--- 9 can be written as 2^3 appprox ... so largest term becomes 2^3 ∗ 2^8 = 2^11
So answer has to be less than 2^11

we have two answer options less than 2^11...how to get into right answer?

That approach might not be applicable for all similar questions. There are two different approaches given in my solution here: 2-99058.html#p764088
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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?  [#permalink]

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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ? &nbs [#permalink] 26 Aug 2018, 04:19
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