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zagcollins
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 Posted: Sun Jul 20, 2008 7:57 am Post subject: Sequences |
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If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ≥ 1, then x20 – x19 =
A.2^19
B.2^20
C.2^21
D.2^20-1
E.2^21-1 |
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pepeprepa
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| Posted: Sun Jul 20, 2008 2:16 pm Post subject: |
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x1=3
x[n+1]=2[xn]-1
x1=3
x2=5
x3=9
You can build this formula which suits to the set: x[n]=(2^n)+1
x[20]=2^20 +1
x[19]=2^19 +1
2^20 + 1 - 2^19 - 1 = 2^19 x (2-1)=2^19
--> A |
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zagcollins
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| Posted: Mon Jul 21, 2008 4:04 am Post subject: |
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| just cant understand the sum...seems tricky and tough.... |
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Ian Stewart
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| Posted: Mon Jul 21, 2008 5:34 am Post subject: Re: Sequences |
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| zagcollins wrote: | If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ? 1, then x20 – x19 =
A.2^19
B.2^20
C.2^21
D.2^20-1
E.2^21-1 |
With a sequence question, it is almost always a good idea to write out the first few terms. Here, we know x(1) = 3. The other terms are defined by the rule:
x(n+1) = 2x(n) - 1
The (n+1)th term is the term that follows the (n)th term, so you can read the above definition as follows: "to find the next term, multiply the current term by 2, then subtract 1". In that way, we can quickly find a few terms:
x(1) = 3
x(2) = 5
x(3) = 9
x(4) = 17
x(5) = 33
What we don't want to do is find the value of x(20) and of x(19)- that could take all day. There are many ways to answer the question, but normally sequence questions test your ability to recognize a pattern. The question asks about the difference between consecutive terms: it asks for x(20) - x(19). We can look at the difference between the terms we wrote down above, and see if a pattern emerges:
x(2) - x(1) = 2 = 2^1
x(3) - x(2) = 4 = 2^2
x(4) - x(3) = 8 = 2^3
x(5) - x(4) = 16 = 2^4
and from this it seems likely that...
x(20) - x(19) = 2^19
That's not a rigorous mathematical proof, but for a GMAT sequence question, it will be a reliable method almost all of the time.
There are many ways to prove this rigorously; one is by 'mathematical induction':
x(1) = 3 = 2^1 + 1
If x(n) = 2^n + 1, then
x(n+1) = 2x(n) - 1 = 2*(2^n + 1) - 1 = 2^(n+1) + 1
So for all n, x(n) = 2^(n) + 1
Thus,
x(20) - x(19) = (2^(20) + 1) - (2^(19) + 1) = 2^20 - 2^19 = 2^19
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pepeprepa
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| Posted: Mon Jul 21, 2008 5:35 am Post subject: |
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x[n] is a term of the set as x[1], x[2],x[3] ....
We know these things:
x[n+1]= 2x[n] -1 for n ≥ 1
x[1]=3
Let's calculate the followinf terms of the set till we understand how the set works.
x1=3
x2=2*3-1=5
x3=2*5-1=9
x4=2*9-1=17
x5=2*17-1=33
...
You calculate the next one till you manage to find that you can find x[n] thanks to this formula x[n]=(2^n)+1
That is just "set practice", do exercices from a maths book and you will get the habit.
Hope it's ok even if i didn't exactly catch what you didn't understand.
If someone has someting clearer (Thanks Ian for your first one method)... |
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zagcollins
Really wants to Beat The GMAT!
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| Posted: Mon Jul 21, 2008 6:56 am Post subject: |
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| thanks Ian and Pepe...great explanation... |
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