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Math Help - another formula...

  1. #1
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    another formula...

    I am not understanding what this question is asking.

    Question: Find another formula suggested by Exercises 3 and 4, and verify your formula by mathematical induction.

    Exercise 3: Prove that 1+3+5+...+(2n-1)=n^2 for every positive integer n by one, by mathematical induction and two, by adding 1+3+5+...+(2n-1) and (2n-1)+(2n-3)+...+1.

    Exercise 4: Use mathematical induction to prove that 1+5+9+...+(4n-3)=2n^2-n for every positive integer n.

    The only idea I have is they want me to change (2n-1) and (4n-3) to fit the series of course.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Possible actuary View Post
    I am not understanding what this question is asking.

    Question: Find another formula suggested by Exercises 3 and 4, and verify your formula by mathematical induction.

    Exercise 3: Prove that 1+3+5+...+(2n-1)=n^2 for every positive integer n by one, by mathematical induction and two, by adding 1+3+5+...+(2n-1) and (2n-1)+(2n-3)+...+1.

    Exercise 4: Use mathematical induction to prove that 1+5+9+...+(4n-3)=2n^2-n for every positive integer n.

    The only idea I have is they want me to change (2n-1) and (4n-3) to fit the series of course.
    So a proof by induction follows a particular pattern (there are slightly different flavors out there, but they amount to the same thing). You have a statement, say P(n) that you want to prove by induction holds for all n. You begin by showing P(1) is true. Then you assume P(k) is true for some k, and show that P(k) => P(k+1)

    This is the most elementary kind of induction proof. they can get more complicated in that you have to proof for numbers higher than 1. I've done induction proofs where i had to prove up to P(6) before the proof could work. anyway, here they are
    Exercise 3:
    Let P(n) be “1 + 3 + 5 + ... + (2n – 1) = n^2” for all positive integer n


    This implies P(1): 2(1) – 1 = 1 = 1^2. So P(1) is true


    Assume P(k) true for some k > 1, we show P(k+1)


    Since P(k) holds, 1 + 3 1 + 3 + 5 + ... + (2k – 1) = k^2
    => P(k+1): 1 + 3 1 + 3 + 5 + ... + (2k – 1) + (2(k+1) – 1)
    .................= P(k) + (2(k+1) – 1)
    .................= k^2 + (2k + 1)
    .................= (k+1)(k+1)
    .................= (k+1)^2
    Thus we see, P(k+1) holds.


    Therefore, we conclude P(n) holds for all n


    Exercise 4:
    Let P(n) be “1+5+9+...+(4n-3)=2n^2-n” for every positive integer n


    This implies P(1): 4(1) – 3 = 1 = 2(1)^2 – 1. So P(1) is true.


    Assume P(k) is true for some k > 1, we show P(k+1)


    Since P(k) holds, 1+5+9+...+(4k-3) = 2k^2 – k
    => P(k+1) : 1 + 5 + 9 + ... + (4k – 3) + (4(k+1) – 3)
    .................= P(k) + (4(k+1) – 3)
    .................= 2k^2 – k + 4(k+1) – 3
    .................= 2k^2 + 4k + 4 – k – 3
    .................= 2k^2 + 4k + 2 – k – 1
    .................= 2(k^2 + 2k + 1) – (k + 1)
    .................= 2(k+1)^2 – (k+1)
    Thus we see, P(k+1) holds.


    Therefore, we conclude P(n) holds for all n.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Exercise 3
    Now this second proof for exercise 3 is a bit sketchy, but I just wanted to give you a general idea of what you should do. Here goes.

    Let P(n) = 1 + 3 + 5 +...+ (2n-1)

    So 2*P(n) = 1 + 3 + 5 +...+ (2n-1) +... + (2n-1) + (2n-3) + ... + 1 .........and let's say 2*P(n) has n terms

    = 2(1) + 2(3) + 2(5) + 2(7) + ...+ 2(2n – 5) + 2(2n – 3) + 2(2n – 1)
    = 2 + 6 + 10 + 14 + ...+ (4n – 10) + (4n – 6) + (4n – 2)
    = [2 + (4n – 2)] + [6 + (4n - 6)] + [10 + (4n - 10)] + [14 + (4n - 14)] + ...
    = 4n + 4n + 4n + 4n + 4n + ....


    notice that to get these 4n's i paired off outer most terms. So there will be half the number of 4n's as there were terms. That is, if we started with n terms we will have (1/2)n terms when we are done pairing off.


    = (1/2)n * 4n .............since the number of 4n's i have will be the number of pairs, which we saw above was (1/2)n
    = 2n^2 = 2*P(n)..........but this the sum of our original series plus itself. So the sum of our original series is half this number

    => P(n) = () * 2n^2 = n^2

    That is, 1+3+5+...+(2n-1) = n^2 as desired
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  4. #4
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    I have done exercise 3 and 4 very similar to what you have done. So, what do they mean when they say "find another formula suggested by exercise 3 and 4"?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Possible actuary View Post
    I have done exercise 3 and 4 very similar to what you have done. So, what do they mean when they say "find another formula suggested by exercise 3 and 4"?
    basically, they want you to come up with another formula for the sums. that is, instead of n^2 or 2n^2 - n, find another formula that calculates the sums based on the patterns you have seen in the exercises. then prove that your formula actually works using induction
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  6. #6
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    Quote Originally Posted by Possible actuary View Post
    I am not understanding what this question is asking.

    Question: Find another formula suggested by Exercises 3 and 4, and verify your formula by mathematical induction.

    Exercise 3: Prove that 1+3+5+...+(2n-1)=n^2 for every positive integer n by one, by mathematical induction and two, by adding 1+3+5+...+(2n-1) and (2n-1)+(2n-3)+...+1.

    Exercise 4: Use mathematical induction to prove that 1+5+9+...+(4n-3)=2n^2-n for every positive integer n.

    The only idea I have is they want me to change (2n-1) and (4n-3) to fit the series of course.
    Try:

    1+6+11+ .. + (5n-4) = (5/2)n^2 - (3/2)n

    RonL
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