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    I have been trying at these for about 6 hours, I have now given up and would be grateful if someone could bring me to the light.

    Using formulas (4) and (5), show that:

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    Formulas (4) and (5):

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    (Original post by Lucid Reality)
    I have been trying at these for about 6 hours, I have now given up and would be grateful if someone could bring me to the light.

    Using formulas (4) and (5), solve these:

    Name:  maths 002.jpg
Views: 91
Size:  309.4 KB



    Formulas (4) and (5):

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    Name:  maths 005.jpg
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    What is the question? To show that the finite* sums on the LHS are equal to the RHS?

    EDIT: sorry finite, not infinite
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    For 8) You have the sum of the squares of odd numbers up to and including 2n+1.
    Observe that this is the same as the sum of the squares of all numbers up to and including 2n+1, minus the sum of the squares of the even numbers up to and including 2n.

    But the sum of the squares of the even numbers up to 2n is simply four times the sum of the squares of ALL numbers up to n! (If you multiply each number from 1 to n by 4 you get the sum of even squares, since the even squares are simply (2n)^2 = 4n^2

    Hence your formula is:

    (The sum of the squares of all numbers up to and including 2n+1) - (The sum of the squares of all EVEN numbers up to and including 2n)

    Which is equal to

    (The sum of the squares of all numbers up to and including 2n+1) - 4*(The sum of the squares of all numbers up to and including n)

    You can apply a similar idea to (9) if you wish. You may also wish to consider using the idea below for finding the general formula for the sum of all squares up to n, and use it to find the general formula for summing all cubes to find the general formula, if you have not been taught it.
    --------------------------------------------------------
    To get the formula for the sum of squares up to n, consider a telescoping sum. Consider the sums

    \sum_{r=0}^n (r+1)^3 - \sum_{r=0}^n (r)^3

    Note the terms cancel out with each other as you add them together, so all you get is (n+1)^3 in the end. Now expand the brackets, and you will get

    \sum_{r=0}^n 3(r)^2 + 3(r) + 1 = (n+1)^3

    How can you use this result to show that:

    \sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)
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    (Original post by Lucid Reality)
    I have been trying at these for about 6 hours, I have now given up and would be grateful if someone could bring me to the light.

    Using formulas (4) and (5), solve these:

    Name:  maths 002.jpg
Views: 91
Size:  309.4 KB



    Formulas (4) and (5):

    Name:  maths 004.jpg
Views: 87
Size:  225.7 KB

    Name:  maths 005.jpg
Views: 72
Size:  253.6 KB
    If you had to prove a result for something like

    1 + 3 + 5 + ... + (2k+1)

    would you know how to tackle that?
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    sorry yeah i should have said "show that.." instead of "solve these"
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    (Original post by CancerousProblem)
    For 8) You have the sum of the squares of odd numbers up to and including 2n+1.
    Observe that this is the same as the sum of the squares of all numbers up to and including 2n+1, minus the sum of the squares of the even numbers up to and including 2n.

    But the sum of the squares of the even numbers up to 2n is simply four times the sum of the squares of ALL numbers up to n! (If you multiply each number from 1 to n by 4 you get the sum of even squares, since the even squares are simply (2n)^2 = 4n^2

    Hence your formula is:

    (The sum of the squares of all numbers up to and including 2n+1) - (The sum of the squares of all EVEN numbers up to and including 2n)

    Which is equal to

    (The sum of the squares of all numbers up to and including 2n+1) - 4*(The sum of the squares of all numbers up to and including n)

    You can apply a similar idea to (9) if you wish. You may also wish to consider using the idea below for finding the general formula for the sum of all squares up to n, and use it to find the general formula for summing all cubes to find the general formula, if you have not been taught it.
    --------------------------------------------------------
    To get the formula for the sum of squares up to n, consider a telescoping sum. Consider the sums

    \sum_{r=1}^n (r+1)^3
    and
    \sum_{r=1}^n (r)^3

    Note the terms cancel out with each other as you add them together, so all you get is (n+1)^3 in the end. Now expand the brackets, and you will get

    \sum_{r=1}^n 3(r)^2 + 3(r) + 1 = (n+1)^3

    How can you use this result to show that:

    \sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)
    Much simpler than that... simply by replacing 'n' with '2n+1' in formula (4) leads easily to the given result for Q8. Use the same logic for the next question.
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    (Original post by lizard54142)
    Much simpler than that... simply by replacing 'n' with '2n+1' in formula (4) leads easily to the given result for Q8. Use the same logic for the next question.
    oh snap yes
    i am a moron
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    (Original post by CancerousProblem)
    oh snap yes
    i am a moron
    That is assuming you are allowed to use formulae (4) and (5) without derivation. Quite a straightforward question once you spot this!
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    (Original post by lizard54142)
    That is assuming you are allowed to use formulae (4) and (5) without derivation. Quite a straightforward question once you spot this!
    i havent got on to that part of the question yet, but yeah if you want to find the sum of powers the best way I can think of to do it is to use a telescoping sum. I'm guessing that's the most efficient way for low powers but for higher powers that would take ages so I don't really know what I would do for that
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    (Original post by Lucid Reality)
    sorry yeah i should have said "show that.." instead of "solve these"
    (Original post by lizard54142)
    Much simpler than that... simply by replacing 'n' with '2n+1' in formula (4) leads easily to the given result for Q8. Use the same logic for the next question.
    OP seeing as it is a "show that" this is your best way to derive these standard results (assuming you can just use the formulae without deriving them). Let us know how it goes
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    Thank you all for your help, I understand Cancer's original way and can solve it using that, but I do not understand this
    (Original post by lizard54142)
    Much simpler than that... simply by replacing 'n' with '2n+1' in formula (4) leads easily to the given result for Q8. Use the same logic for the next question.
    ?

    so say Gn = 1^2 + 2^2 + ... + n^2

    simply replacing n with 2n+1 would yield 2n+1 elements within the series, and would not take care of removing the even squares? i.e.

    G(2n+1) = 1^2 + 2^2 + ... + (2n + 1)^2
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    (Original post by Lucid Reality)
    Thank you all for your help, I understand Cancer's original way and can solve it using that, but I do not understand this

    ?

    so say Gn = 1^2 + 2^2 + ... + n^2

    simply replacing n with 2n+1 would yield 2n+1 elements within the series, and would not take care of removing the even squares? i.e.

    G(2n+1) = 1^2 + 2^2 + ... + (2n + 1)^2
    Okay, I will go through the initial part of question 8 with you.

    

1^2 + 2^2 + 3^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}

for k \in \mathbb{N}



let 2n+1 = k for n \in \mathbb{N}

...

    Can you see where to go from here?
 
 
 

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