If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
SOLUTIONS/COMMENTARY ON QUERY 16
**** Query 11.1.6 first five terms of { (-1)^(n-1) (n/(2n-1))
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Your solution:
Confidence Assessment:
Given Solution:
** Subsituting we obtain
a(1) = (-1)^(1-1) * 1 / (2*1 - 1) = (-1)^0 * 1 / 1 = 1
a(2) = (-1)^(2-1) * 2 / (2*2 - 1) = (-1)^1 * 2 / 3 = -2/3
a(3) = (-1)^(3-1) * 3 / (2*3 - 1) = (-1)^2 * 3 / 5 = 3/5
a(4) = (-1)^(4-1) * 4 / (2*4 - 1) = (-1)^3 * 4 / 7 = -4/7
a(5) = (-1)^(5-1) * 5 / (2*5 - 1) = (-1)^4 * 5 / 9 = 5/9 **
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Query 11.1.16 nth term of 2/3,4/9,8/27, 16/81,...
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Your solution:
Confidence Assessment:
Given Solution:
The nth term of 2/3,4/9,8/27, 16/81,... is
a(n) = { (2^n) / (3^n) }
** Good. Note that (2^n) / (3^n) = (2/3)^n.
The sequence consists of powers of 2/3.
We can therefore write the sequence as a(n) = (2/3)^n. **
Self-critique (if necessary):
Self-critique Rating:
**** Query 11.1.18 nth term of 1, 1/2, 3/, 1/4, ...
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Your solution:
Confidence Assessment:
Given Solution:
a sub n= 1/n if n is even
a sub n = n if n is odd
** Good, but the even-odd thing requires a decision on the part of the user. Better to find a single expression to define the sequence.
If you raise n to the +1 power when n is odd and to the -1 power when n is even, you get the correct sequence.
So we want to find an expression that's 1 whenever n is odd and -1 whenever n is even. Noting that the powers of -1 alternate between 1 and -1, we could try (-1)^n. However that gives -1 when n is odd and 1 when n is even. So we try (-1)^(n-1), which does give us 1 when n is odd and -1 when n is even.
So we use this expression for our exponent, and get the single expression that defines the sequence: n^( -1^(n-1)). **
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**** Query 11.1.30 a1 = -1, a2 = 1, a(n+2) = a(n+1) + n an.
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Your solution:
Confidence Assessment:
Given Solution:
** For n = 1 we get a(1+2) = a(1+1) + 1 * a(1), or a(3) = a(2) + 1 * a(1).
This gives us a(3) = 1 + 1 * -1 = 1 - 1 = 0.
Then a(4) = a(3) + 2 a(2) = 0 + 2 * 1 = 2.
Then a(5) = a(4) + 3 a(3) = 2 + 3 * 0 = 2.
Then a(6) = a(5) + 4 a(4) = 2 + 4 * 2 = 10. **
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**** Query 11.1.40 sum{(3k - 7), k, 1,6)
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14:37:46
Your solution:
Confidence Assessment:
Given Solution:
** We are to sum the first six terms of the sequence.
We calculate:
a(1) = ((3(1)-7)=-4
a(2) = ((3)(2)-7=-1
a(3) = ((3)(3)-7)=2
a(4) = ((3)(4)-7)=5
a(5) = ((3)(5)-7)=8
a(6) = ((3)(6)-7)=11
Our total is a(1) + a(2) + a(3) + a(4) + a(5) + a(6) = 21. **
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**** Query 11.1.50 write out sum( (k+1)^2, k, 1, n).
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Your solution:
Confidence Assessment:
Given Solution:
** Our terms are
a(1) = (1+1)^2=4
a(2) = (2+1)^2=9
a(3) = (3+1)^2=16
a(4) = (4+1)^2=25
a(5) = (5+1)^2=36
. . .
a(n) = (n+1)^2
The sum is therefore written as
4+9+16+25+36+49...+ (n+1)^2 **
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**** Query 11.1.60 express using summation notation 1 + 3 + ... +
[2(12)-1]
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14:44:02
Your solution:
Confidence Assessment:
Given Solution:
*&*& Using summation notation .
we would have at the top of the sigma the letter 12 to represent how many numbers the sequence goes through.
at the bottom of sigma we have k=1 and to the right of it the equation 2k-1
In DERIVE notation the expression is sum(2k-1, k, 1, 12), indicating the sum of the expression 2k-1 from k = 1 to k = 12. *&*&
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**** Query 11.1.72 (5th ed #68) un = [ (1 + `sqrt(5))^n - (1 -
`sqrt(5)^n ] / (2^n `sqrt(5) )
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14:50:43
Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION
for part
a) we can say that u sub 1 equals 1 because in the Fibonacci sequence u
sub 1 and u sub 2 are equal. and the third term of the sequence is the sum
of u sub 1 and u sub 2.
b) in the sequence each new number is the addition of the current number and
the starting number which is in this case u sub n. so u sub n+1 + u sub n= u
sub n+2
c)we know that u sub n is the Fibonacci sequence for various reasons. the
main reason is the fact that if both equations in the numerator are the same
and the exponents are the same then those equations will according to the
sequence will be an even number which can be divided by the denominator.
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14:50:43
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Self-critique (if necessary):
Self-critique Rating:
**** Explain how you showed that u1 = 1 and u2 = 1.
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20:23:48
Your solution:
Confidence Assessment:
Given Solution:
*&*&
For u1 you let n = 1 and for u2 you let n = 2:
u1 = [ (1 + `sqrt(5))^1 - (1 - `sqrt(5)^1 ] / (2^1`sqrt(5) )
= ( 1 + 'sqrt(5) -1 + 'sqrt(5) ) / 2 'sqrt(5)
= ( 2 'sqrt(5) ) / ( 2 'sqrt(5) )
= 1
and
u2 = [ (1 + `sqrt(5))^2 - (1 - `sqrt(5)^2 ] / (2^2`sqrt(5) )
= ( 1 + 'sqrt(5) + 'sqrt(5) + 5 ) - ( 1 -'sqrt(5) -'sqrt(5) + 5 ) / 4 'sqrt(5)
= ( 4 'sqrt(5) ) / ( 4 'sqrt(5) )
= 1
*&*&
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20:23:49
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Self-critique Rating:
**** Explain how you showed that u(n+2) = u(n+1) + un.
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20:28:53
Your solution:
Confidence Assessment:
Given Solution:
After workin out u1 and u2, which equalled 1 and u3 , which equalled 2.
I then aworked u3 which is u2 plus u1 to get 1 + 1 = 2
u4 = u3 + u2 = 2 + 1 = 3
Because obtaineing the third term required the 1st and 2nd terms ( and so forth) this is the Fibonacci sequence.
** To show this for general n you have to simplify the expression for u(n+1) + un to show that it's the same as that for u(n+2).
[ (1 + `sqrt(5))^(n+2) - (1 - `sqrt(5))^(n+2) ] / (2^(n+2) `sqrt(5) ) =
[ (1 + `sqrt(5))^(n+1) - (1 - `sqrt(5))^(n+1) ] / (2^(n+1) `sqrt(5) )
+ [ (1 + `sqrt(5))^n - (1 - `sqrt(5)^n ] / (2^n `sqrt(5) ).
You begin by multiplying both sides 2^n * `sqrt(5), which eliminates part of the denominator. They multiply both sides by the common denominator 4 of the remaining expressions.
Then use the distributive law, being careful, expanding everything into individual terms then show that everything just cancels out, leaving you 0 = 0.**
for the Fibonacci sequence we must be able to say u sub 3= u sub 1 + u sub 2
which in this case would equal 2. then we can move up to u sub 4 which would
be u sub 2+ u sub 3 which = 3 and u sub 5= u sub 3 + u sub 4= 5 and this
equation fits that pattern.
** You obtained your u1, u2, u3 ... values by referring to the Fibonacci sequence.
Then you used this to justify your statement that the expression represents the Fibonacci sequence.
So you are saying that the sequence represents the Fib seq because the expression represents the Fib seq.
That's circular.
If you had algebraically evaluated the expression for n = 1, n = 2, etc., and shown that the numbers are indeed identical to those of the Fibonacci sequence, then you wouldn't have proven the result but you would have a good indication that it works. **
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14:51:32
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Self-critique (if necessary):
Self-critique Rating:
**** Query 11.2.18 nth term of arithmetic sequence with a = 0, d =
`pi.
**** What is your expression for the nth term of the sequence?
Your solution:
Confidence Assessment:
Given Solution:
NOT-QUITE-CORRECT STUDENT SOLUTION
a sub n= pi(n)-n
(pi)(n)-pi
** Term n+1 would then be `pi * (n+1) - (n+1), and the difference between the nth term and the n+1 term would be
`pi * (n+1) - (n+1) - [ `pi * n - n] = `pi * n+ `pi * 1 - `pi * n - n - 1 + n = `pi - 1.
But the difference should be `pi.
The correct expression for the nth term is a(n) = `pi * n. **
The sequence begins with a(0) = 0 and each term adds pi to the preceding term.
So
a(1) = a(0) + pi = 0 + pi = pi
a(2) = a(1) + pi = pi + pi = 2 pi
a(3) = a(2) + pi = 2 pi + pi = 3 pi
etc.
At the nth step we get
a(n) = n pi.
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14:52:32
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**** Query 11.2.24 7th term of 2 `sqrt(5), 4 `sqrt(5), 6 `sqrt(5), ...
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14:53:01
Your solution:
Confidence Assessment:
Given Solution:
Not-quite-correct student solution:
a sub 7= 14(sq.rt.5)
a7 = 2 'sqrt(5) + 12
Here's how I got it.
I used a(n) = a + (n-1)d ; a = 2 'sqrt(5) ; d = 2
a7 = 2 'sqrt (5) + (7-1) 2
= 2 'sqrt (5) + 12
** If the sequence is as I stated above, the common difference is 2 `sqrt(5) and the 7th term is 14 `sqrt(5). **
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14:53:02
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14:54:20
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**** Query 11.2.30 5th term -2, 13th term 30. ......!!!!!!!!...................................
14:58:53
Your solution:
Confidence Assessment:
Given Solution:
If a(n) = a + n * d then we have
a(5) = a + 5 * d = -2 and
a(13) = a + 13 * d = 30 .
Subtracting
a + 5 * d = -2 from
a + 13 * d = 30 we get
8 * d = 32 so that d = 4.
Substituting this into
a + 5 * d = -2 we get
a + 5 * 4 = -2 so that
a = -2 – 5 * 4 = -22.
Our expression for a(n) is therefore
a(n) = -22 + 4 * n.
we can find the value for d and the first number in the sequence by saying
a sub 5=a +4d=-2
a sub 13=a +12d=30
by subtracting the second equation from the first we get -8d=-32 so d=4 we
can then find the value of a sub 1 by plugging in the value of d=4 to the
equation a sub 5. so a= -2- 4d so a=-2-(4)(4)
so a sub 1=-18
the formula we could use is
a sub n=a +(4n-4)
** If d = 4 and a(1) = -18 then we have a(n) = a(n-1) + 4, with a(1) = -18. **
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14:58:57
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**** Query 11.2.36 sum of -1+3+7+...+(4n-5).
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15:03:32
Your solution:
Confidence Assessment:
Given Solution:
by using the equation S sub n = n/2(a+a sub n) we get
s sub n= n/2(-1+(4n-5)= (n/2)(4n-6)
** The sum of -1 and (4n-5) is 4n - 6. If we pair the numbers in the usual fashion this is the sum of every pair.
We have n/2 pairs so the sum is n/2 ( 4n - 6) = 2 n^2 - 3 n. **
** Good. This simplifies to give us n * ( 2n - 3 ). If expanded form is desired you would have 2 n^2 - 3 n. **
Using the formula Sn = n/2 (a + a(n) ) I got...
Sn = n /2 ( 8 + (4n - 5 ))
= n/2(4n +3)
** Shouldn't that be Sn = n/2 ( a(1) + a(n) ) =
n/2 (-1 + (4n-5)) =
n/2 (4n - 6) =
2 n^2 - 3 n? **
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15:03:40
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**** Query 11.2.52 staircase 30 steps, bottom requires 100 bricks,
each successive step 2 less than previous.
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15:10:58
Your solution:
Confidence Assessment:
Given Solution:
** Step n requires a0 - 2 n bricks.
Since step 1 requires 100 bricks, a1 = a0 - 2 * 1 = 100. Solving for a0 we get a0 = 102.
So step n requires 102 - 2 * n bricks. That is, an = 102 - 2 * n.
It follows that a30 = 102 - 2 * 30 = 42.
The total number of brick is
number of pairs * total for each pair
= n/2 ( a1 + an)
= 30/2 ( 100 + 42)
= 15 * 142
= 2130. **
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