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2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?
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12 Dec 2022, 06:17
12 Dec 2022, 06:17
Expert Reply
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Question Stats:
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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. \(2^{37}\)
Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?
[#permalink]
14 Dec 2022, 04:33
14 Dec 2022, 04:33
1
Expert Reply
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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Re: 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?
[#permalink]
14 Jan 2023, 23:12
14 Jan 2023, 23:12
1
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
