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    could i have some help with edexcel FP1, ex6A question 10. I have started of by doing the n=1 case, and then assume n=k for k=all positive integers...

    then i am confused when it comes to substituting in the k+1 for 2n, and also how to finish off the proof, because i can't seem to re-arrange mine to get back to the original K+1 formula
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    It is easier if you post the question
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    Here's the question.

    Prove by mathematical induction for n \in Z^+

    \displaystyle\sum_{r=1}^{2n}r^2 = \frac{1}{3}n(2n+1)(4n+1)
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    When you replace k with k + 1;
    On the left the sum goes up to 2k + 2
    which means you need to include two new terms on the right...
    (2k + 1)2 and (2k + 2)2

    And then you have to show that the rhs including the new terms is equivalent to

    1/3( k + 1 )( 2k + 3 )( 4k + 5 )

    which you can do by expanding and comparing rather than factorising
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    (Original post by SubAtomic)
    Here's the question.

    Prove by mathematical induction for n \in Z^+

    \displaystyle\sum_{r=1}^{2n}r^2 = \frac{1}{3}n(2n+1)(4n+1)
    Use n for the proof
    Let the sum be true for n
    that is
    \displaystyle \sum_{r=1}^n r^2=\frac{n}{6}(n+1)(2n+1)
    for n+1
    \displaystyle \sum_{r=1}^{n+1}r^2=\sum_{r=1}^n r^2+(n+1)^2=
    \displaystyle =\frac{n}{6}(n+1)(2n+1)+(n+1)^2=  \frac{n+1}{6}[n(2n+1)+6(n+1)]
    Factorize the terms in the brackets with f.e the quadratic formula
    an^2+bn+c=a(n-n_1)(n-n_2)=(an-an_1)(n-n_2)
    where n_1 and n_2 is the two roots of the
    quadratic equation.
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    I understand the method, but could you please explain, why I have to add two new terms, (2k + 1)2 and (2k + 2)2
    but could you explain where the (2k + 1)^2 term comes from. I thought that the only extra term was (2k + 2)^2.
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    (Original post by Jaisal Patel)
    I understand the method, but could you please explain, why I have to add two new terms,
    Your sum to k is \displaystyle\sum_{r=1}^{2k}r^2

    then your sum to k+1, which comes from replacing k with k+1, is \displaystyle\sum_{r=1}^{2(k+1)}  r^2=\sum_{r=1}^{2k+2}r^2

    hence two additional terms.
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    thanks,
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    (Original post by ztibor)
    Use n for the proof
    Let the sum be true for n
    that is
    \displaystyle \sum_{r=1}^n r^2=\frac{n}{6}(n+1)(2n+1)
    for n+1
    \displaystyle \sum_{r=1}^{n+1}r^2=\sum_{r=1}^n r^2+(n+1)^2=
    \displaystyle =\frac{n}{6}(n+1)(2n+1)+(n+1)^2=  \frac{n+1}{6}[n(2n+1)+6(n+1)]
    Factorize the terms in the brackets with f.e the quadratic formula
    an^2+bn+c=a(n-n_1)(n-n_2)=(an-an_1)(n-n_2)
    where n_1 and n_2 is the two roots of the
    quadratic equation.
    I've always being curious to these FP1 questions, what is the method of algebraic manipulation used here \displaystyle =\frac{n}{6}(n+1)(2n+1)+(n+1)^2=  \frac{n+1}{6}[n(2n+1)+6(n+1)] Is this instantaneous, or are there more steps, what exactly has happened here, as with every proof by induction question its not the context and instructions that throw me off, its the algebraic manipulation.
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    (Original post by Robbie242)
    I've always being curious to these FP1 questions, what is the method of algebraic manipulation used here \displaystyle =\frac{n}{6}(n+1)(2n+1)+(n+1)^2=  \frac{n+1}{6}[n(2n+1)+6(n+1)] Is this instantaneous, or are there more steps, what exactly has happened here, as with every proof by induction question its not the context and instructions that throw me off, its the algebraic manipulation.
    It is instant, it's just factoring out \dfrac{n+1}{6} from both terms.
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    (Original post by Noble.)
    It is instant, it's just factoring out \dfrac{n+1}{6} from both terms.
    And for most induction questions would you use this method of factorisation (i.e the series ones) I can see where the terms come from now though
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    (Original post by Robbie242)
    I've always being curious to these FP1 questions, what is the method of algebraic manipulation used here \displaystyle =\frac{n}{6}(n+1)(2n+1)+(n+1)^2=  \frac{n+1}{6}[n(2n+1)+6(n+1)] Is this instantaneous, or are there more steps, what exactly has happened here, as with every proof by induction question its not the context and instructions that throw me off, its the algebraic manipulation.
    It was a factoring out as Noble. wrote
    And arranging after that
    \displaystyle \frac{n+1}{6}\left (2n^2+7n+6\right )
    the two roots of quadratic equation
    \displaystyle n_{1,2}=\frac{-7 \pm \sqrt{49-48}}{4}=\frac{-7 \pm 1}{4}
    that is \displaystyle n_1=\frac{-3}{2} and n_2=-2
    so the term will be
    \displaystyle \frac{n+1}{6}\left [2\left(n+2\right )\left (n+\frac{3}{2}\right )\right]=
    \displaystyle \frac{n+1}{6}\left (n+2\right )\left (2n+3\right)=
    \displaystyle =\frac{n+1}{6}\left [ \left (n+1\right )+1\right ]\left [2\left (n+1\right)+1\right ]
    Which is the formula for n+1 and the proof is completed.
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    (Original post by ztibor)
    It was a factoring out as Noble. wrote
    And arranging after that
    \displaystyle \frac{n+1}{6}\left (2n^2+7n+6\right )
    the two roots of quadratic equation
    \displaystyle n_{1,2}=\frac{-7 \pm \sqrt{49-48}}{4}=\frac{-7 \pm 1}{4}
    that is \displaystyle n_1=\frac{-3}{2} and n_2=-2
    so the term will be
    \displaystyle \frac{n+1}{6}\left [2\left(n+2\right )\left (n+\frac{3}{2}\right )\right]=
    \displaystyle \frac{n+1}{6}\left (n+2\right )\left (2n+3\right)=
    \displaystyle =\frac{n+1}{6}\left [ \left (n+1\right )+1\right ]\left [2\left (n+1\right)+1\right ]
    Which is the formula for n+1 and the proof is completed.
    Wow thanks so much, I never knew the quadratic step appreciate the alternate method!
 
 
 
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