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18 Jun 2017, 06:41
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Difficulty:
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Question Stats:
76% (01:29) correct
24%
(01:38)
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\(4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 =\)
A. 2^7
B. 2^8
C. 2^16
D. 2^28
E. 2^29
A. 2^7
B. 2^8
C. 2^16
D. 2^28
E. 2^29
18 Jun 2017, 06:58
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Bunuel
\(4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 =\)
A. 2^7
B. 2^8
C. 2^16
D. 2^28
E. 2^29
A. 2^7
B. 2^8
C. 2^16
D. 2^28
E. 2^29
Look for a pattern
4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = 4 + 4 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7
= 8 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7
= 2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7
Now examine the following...
2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = 8 + 8 + 2^4 + 2^5 + 2^6 + 2^7
= 16 + 2^4 + 2^5 + 2^6 + 2^7
= 2^4 + 2^4 + 2^5 + 2^6 + 2^7
And then...
2^4 + 2^4 + 2^5 + 2^6 + 2^7 = 16 + 16 + 2^5 + 2^6 + 2^7
= 32 + 2^5 + 2^6 + 2^7
= 2^5 + 2^5 + 2^6 + 2^7
In general, 2^n + 2^n = 2^(n+1)
This should make sense since 2^n + 2^n = 2^n(1 + 1) = (2^n)(2^1) = 2^(n+1)
So, continuing with the pattern, we get a TOTAL SUM of 2^8
Answer:
General Discussion
sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
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Location: India
Concentration: Technology, Strategy
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
WE:Information Technology (Consulting)
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 612
Bunuel
\(4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 =\)
A. 2^7
B. 2^8
C. 2^16
D. 2^28
E. 2^29
A. 2^7
B. 2^8
C. 2^16
D. 2^28
E. 2^29
\(4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7\)
\(4 = 2^2\), therefore;
\(2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7\)
\(2^2(1 + 1+2+2^2+2^3+2^4 + 2^5)\)
\(2^2(4+4+8+16+32)\)
\(2^2(64)\) ------------- (\(64 = 2^6\))
\(2^2(2^6)\) \(= 2^{2+6}\)
\(2^8\) . Answer (B)...
