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exponents

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by scoobydooby » Sat May 16, 2009 8:09 am
2+2+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+[sum of GP with common ratio 2, n=8; a(r^n-1)/(r-1)]
=2+2(2^8-1)/(2-1)
=2+2(2^8-1)
=2+2^9-2
=2^9
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by dumb.doofus » Sat May 16, 2009 8:10 am
You just have to apply geometric series here..

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

= 2 + 2 (1+2+2^2+2^3+2^4+2^5+2^6+2^7)

geometric series = a(r^n - 1)/(r-1)
here a = 1, r = 2, n = 8, applying it above gives us

= 2 + 2(2^8 -1)

= 2^9
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Re: exponents

by dtweah » Sat May 16, 2009 8:13 am
kevind147 wrote:from a practice test:

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

I got e., but the answer is a. Why?
See the pattern:
2+2=4=2^2
2^2+2^2=8=2^3

So continuing this way and inducting on the observed pattern, you will end up with

2^8 +2^8=2^9.
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by kevind147 » Sat May 16, 2009 8:18 am
scoobydooby wrote:2+2+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+[sum of GP with common ratio 2, n=8; a(r^n-1)/(r-1)]
=2+2(2^8-1)/(2-1)
=2+2(2^8-1)
=2+2^9-2
=2^9
thanks this makes the most sense to me, but why is the series multiplied by 2? Why is it 2(2^8-1) instead of just (2^8-1)?
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by scoobydooby » Sat May 16, 2009 8:29 am
that comes from the formula a(r^n-1)/(r-1)

our series: 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8
a=2, r=2, n=8
putting it all in the formula gives 2(2^8-1)
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Re: exponents

by dumb.doofus » Sat May 16, 2009 10:29 am
dtweah wrote:
kevind147 wrote:from a practice test:

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

I got e., but the answer is a. Why?
See the pattern:
2+2=4=2^2
2^2+2^2=8=2^3

So continuing this way and inducting on the observed pattern, you will end up with

2^8 +2^8=2^9.
Nice approach.. quite simple and fast..
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