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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. \(2^{37}\)
A. \(2^9\)
B. \(2^{10}\)
C. \(2^{16}\)
D. \(2^{35}\)
E. \(2^{37}\)
Solving this question helps.
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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
\(2+2=2*2=2^2\) (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.
Answer: A.
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:
\(2+(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)=2+\frac{2*(2^{8}-1)}{2-1}=2^9\).
Answer: A.
_________________
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.
Answer: A.
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:
\(2+(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)=2+\frac{2*(2^{8}-1)}{2-1}=2^9\).
Answer: A.
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Posts: 149
there are nine numbers, so group them up in 3.
\([2+2+2^2]+[2^3+2^4+2^5]+[2^6+2^7+2^8]\)
\([2^3]+2^3[1+2+4]+2^6[1+2+4]=2^3+2^3*7+2^3*8*7\)
\(2^3*64=2^3*2^6=2^9\)
\([2+2+2^2]+[2^3+2^4+2^5]+[2^6+2^7+2^8]\)
\([2^3]+2^3[1+2+4]+2^6[1+2+4]=2^3+2^3*7+2^3*8*7\)
\(2^3*64=2^3*2^6=2^9\)
General Discussion
Posts: 148
amitdgr
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..!!
cangetgmat
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 = \(1 + 2^9 - 1 = 2^9\)
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Check similar questions to practice:
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tough-and-tricky-exponents-and-roots-questions-125956-40.html?hilit=geometric#p1029222
sequence-s-is-defined-as-follows-s1-2-s2-2-1-s3-2-2-sn-129435.html
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3-108690.html
tough-and-tricky-exponents-and-roots-questions-125956-40.html?hilit=geometric#p1029222
sequence-s-is-defined-as-follows-s1-2-s2-2-1-s3-2-2-sn-129435.html
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Baten80
don't neglect what might be the easiest way to solve this problem -- and what's absolutely the most straightforward: just convert these to concrete numbers.
the problem statement is
2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256
the first few terms here are pretty insignificant, as far as the total sum is concerned. 32 + 64 is pretty close to 100, so, this sum will be close to 100 + 128 + 256.
the answer choices are
A/ 512
B/ 1024
C/ ha, you must be kidding
D/ ha, you must be kidding
E/ ha, you must be kidding.
this method certainly doesn't take any MORE time than any of the other methods here... and it may in fact take less, for many people.
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Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 417
Location: United Kingdom
GMAT 1: 450 Q42 V34
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GMAT 1: 450 Q42 V34
Posts: 417
kraizada84
Great explanation kraizada84, one question why is 2^2+2^2=2^3 and not 2^4 ? Same with 2^3 + 2^3=2^4 and not 2^6?....
Thanks for your time in advanced.
Kimberly77
Recommend studying up on exponents.
2^2 + 2^2 = 2^2(1+1) = 2^2(2)= 2^3
Exponents are added only when two like terms are multiplied
2^2 * 2^2 = 2^4
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Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 417
Location: United Kingdom
GMAT 1: 450 Q42 V34
Products:
GMAT 1: 450 Q42 V34
Posts: 417
Great thanks Regor60. My silly mistake and make sense now.
Solving using two methods
Method 1: Logic
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
Following the same pattern till last term we will get
.
.
.
= \(2^8 + 2^8\)
= \(2*2^8\)
= \(2^9\)
So, Answer will be A
Method 2: Geometric series formula
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= 2 + a Geometric series whose first term is 2, common ratio is 2 and number of terms is 8
= 2 + 2*\(\frac{2^8 - 1 }{ 2 - 1}\)
= 2 + \(2^9\) - 2
= \(2^9\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Sequence problems
_________________
Method 1: Logic
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
Following the same pattern till last term we will get
.
.
.
= \(2^8 + 2^8\)
= \(2*2^8\)
= \(2^9\)
So, Answer will be A
Method 2: Geometric series formula
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= 2 + a Geometric series whose first term is 2, common ratio is 2 and number of terms is 8
= 2 + 2*\(\frac{2^8 - 1 }{ 2 - 1}\)
= 2 + \(2^9\) - 2
= \(2^9\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Sequence problems
_________________
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GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
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GMAT Focus 1: 575 Q77 V81 DI78
Posts: 106
All the methods are great but counting them in my head was the fastest for me lol
GMAT 1: 750 Q49 V42
Posts: 469
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