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sev Just gettin' started!
Joined: 06 Apr 2007 Posts: 26
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Posted: Fri Jun 08, 2007 5:03 pm Post subject: Adding exponents problem from pract. Test |
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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
Obviously it's not E, but how do you do this problem? The OA is A which is 2^9
Is there any way to do it without actually calculating all of the values, cuz that just doesnt seem right to me. |
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smashingdon Just gettin' started!
Joined: 29 May 2007 Posts: 2
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Location: Pune, India
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Posted: Mon Jun 11, 2007 5:10 am Post subject: its not that difficult |
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even if you work it out its simple.
expand just the first few ones
it will become
2 + 2 + 4 + 8 + 16 + 2^5 + 2^6 + 2^7 + 2^8
will become
32 + 2^5 + 2^6 + 2^7 + 2^8
2^5 + 2^5 + 2^6 + 2^7 + 2^8
2^5[1+1+2+4+8]
2^5 * 16
2^5 * 2^4
2^9 |
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bills_e Just gettin' started!
Joined: 12 Mar 2007 Posts: 1
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Posted: Tue Jul 17, 2007 3:55 pm Post subject: Re: Adding exponents problem from pract. Test |
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| sev 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
Obviously it's not E, but how do you do this problem? The OA is A which is 2^9
Is there any way to do it without actually calculating all of the values, cuz that just doesnt seem right to me. |
2+2 = 2^2. (Adding the first two terms)
2^2 + 2^2 = 2^3 (Adding the above result to the third term)
2^3 + 2^3 = 2^4 (Adding the above result to the fourth term)
.....
2^8+2^8 = 2^9
Therefore, A. |
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