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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. \(2^{37}\)
14 Aug 2010, 10:18
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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.
01 Apr 2012, 14:24
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there are nine numbers, so group them up in 3.
\([2+2+2^2]+[2^3+2^4+2^5]+[2^6+2^7+2^8]\)
\([2^3]+2^3[1+2+4]+2^6[1+2+4]=2^3+2^3*7+2^3*8*7\)
\(2^3*64=2^3*2^6=2^9\)
\([2+2+2^2]+[2^3+2^4+2^5]+[2^6+2^7+2^8]\)
\([2^3]+2^3[1+2+4]+2^6[1+2+4]=2^3+2^3*7+2^3*8*7\)
\(2^3*64=2^3*2^6=2^9\)
