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    Prove that

    \displaystyle \sum_{i=1}^{n} (i+1)^3 - \sum_{i=1}^{n} i^{3} = (n+1)^{3}-1

    Here's my working out starting from the LHS,

    \displaystyle \sum_{i=1}^{n} (i^{3}+3i^{2}+3i+1) - \sum_{i=1}^{n} i^{3}

    \displaystyle \sum_{i=1}^{n} i^{3} + 3 \sum_{i=1}^{n} i^{2} + 3\sum_{i=1}^{n} i + \sum_{i=1}^{n} 1 - \sum_{i=1}^{n} i^{3}

    \displaystyle 3 \sum_{i=1}^{n} i^{2} + 3\sum_{i=1}^{n} i + \sum_{i=1}^{n} 1

    Using the summation results for i and i^{2}

    \displaystyle 3(\frac{n}{6}(2n+1)(n+1))+3( \frac{n}{2} (n+2))+n

    \displaystyle \frac{n}{2}[(2n^{2}+3n+1)+(3n+6)+2n]

    Thus obtaining  \frac{n}{2}(2n^{2}+8n+7)

    Where am I going wrong?

    Yes I can use Proof by Induction but I'm taking the approach that'll lead me to (n+1)^{3}-1 thus proving LHS = RHS.

    Any help is appreciated
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    You factored wrong:

    n/2 [(n+1)(2n+1) + 3(n+1) + 2]

    Notice the 2. You made a mistake there.

    n/(n/2) = 2, not 2n.
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    should be 2, not 2n in the square brackets.
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    (Original post by thekidwhogames)
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    Ahh right, I see what you mean, I didn't see the two mistakes.

    Thank you very much!
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    (Original post by ManLike007)
    Ahh right, I see what you mean, I didn't see the two mistakes.

    Thank you very much!
    No problem! By the way, is this FP1 revision?
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    (Original post by the bear)
    should be 2, not 2n in the square brackets.
    Oh damn I didn't see your comment, sorry! But besides that, I appreciate your contribution!
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    (Original post by thekidwhogames)
    No problem! By the way, is this FP1 revision?
    I wish but no it's a Series and Sequences topic for my uni module. I'm doing a problem solving sheet which starts off easy like the post above but later on it gets funky with limits, diverging & converging sequences, L'Hopital's Rule and all of that good stuff!
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    (Original post by ManLike007)
    I wish but no it's a Series and Sequences topic for my uni module. I'm doing a problem solving sheet which starts off easy like the post above but later on it gets funky with limits, diverging & converging sequences, L'Hopital's Rule and all of that good stuff!
    Ahh, like Calculus 2? Convergent sequences, integrals test and L'hopital? Good stuff indeed!
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    (Original post by ManLike007)
    Prove that

    \displaystyle \sum_{i=1}^{n} (i+1)^3 - \sum_{i=1}^{n} i^{3} = (n+1)^{3}-1
    To my mind, the obvious thing to do here is observe that reindexing the first sum (with j=i+1) gives \displaystyle \sum_{i=1}^{n} (i+1)^3 = \sum_{j=2}^{n+1} j^3 and the result is immediate.

    I can understand feeling happier *justifying* the result by multiplying out as you've done, but to be honest it's much more important that you understand the reindexing approach.
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    (Original post by DFranklin)
    To my mind, the obvious thing to do here is observe that reindexing the first sum (with j=i+1) gives \displaystyle \sum_{i=1}^{n} (i+1)^3 = \sum_{j=2}^{n+1} j^3 and the result is immediate.
    Firstly, thank you very much for your input!

    Secondly, I understand how you 'reindexed' it but I'm sorry if I sound silly but why does this become obvious? What is even the purpose of reindexing the sum?
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    (Original post by ManLike007)
    Firstly, thank you very much for your input!

    Secondly, I understand how you 'reindexed' it but I'm sorry if I sound silly but why does this become obvious? What is even the purpose of reindexing the sum?
    So, after reindexing, you're trying to find

    \displaystyle \sum_{j=2}^{n+1} j^3 - \sum_{i=1}^n i^3

    Nearly everything is going to cancel, leaving the result.

    If you can't see this, write out the terms in each sum for, say, n = 5 and it should be obvious how it works.

    This is one of those things where there isn't really very good mathematical language to explain what's going on, but if you understand what summation is, it's actually very straightforward. I'd expect someone with that understanding to instantly see that the difference between the sums is (n+1)^3 - 1 (the difficulty then being explaining *why* that is the case).
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    (Original post by DFranklin)
    If you can't see this, write out the terms in each sum for, say, n = 5 and it should be obvious how it works.
    (Original post by DFranklin)
    instantly see that the difference between the sums is (n+1)^3 - 1
    Ohh ok I see what you mean, that's quite interesting
 
 
 
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