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Thread: Proof by mathematical induction

  1. #1
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    Proof by mathematical induction

    I want to prove

    $\displaystyle (n + 1)^2 + (n + 2)^2 + (n + 3)^2 + \ldots + (2n)^2 = \frac{n(2n + 1)(7n + 1)}{6}, \forall n \geq 1$

    or

    $\displaystyle \sum_{i=1}^{n} (n + i)^2 = \frac{n(2n + 1)(7n + 1)}{6}, \forall n \geq 1$

    by mathematical induction.

    I tried with n = 1 and I got

    Left side: $\displaystyle (1 + 1)^2 = 2^2 = 4$
    Right side: $\displaystyle \frac{1(2 + 1)(7 + 1)}{6} = \frac{3 * 8}{6} = 4$

    4 = 4, so I assume $\displaystyle k = n - 1 \implies k^2 + (k + 1)^2 + (k + 2)^2 + \ldots + (2(k-1))^2 = \frac{k(2k + 1)(7k + 1)}{6}, \forall k > 1$

    Now I want to prove it for $\displaystyle n = k$ but I don't even know how to start... What should I do? Thanks!
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  2. #2
    MHF Contributor
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    Re: Proof by mathematical induction

    we know that

    $\sum \limits_{i=1}^n~i^2 = \dfrac{n(n+1)(2n+1)}{6}$

    so

    $\sum \limits_{i=1}^{2n}~i^2 = \dfrac{2n(2n+1)(4n+1)}{6}$

    $\sum \limits_{n+1}^{2n}~i^2 =\dfrac{2n(2n+1)(4n+1)}{6}-\dfrac{n(n+1)(2n+1)}{6} =$

    $\dfrac 1 6 \left(16 n^3+12 n^2+2 n - (2 n^3+3 n^2+n)\right) = $

    $\dfrac 1 6 (n + 9 n^2 + 14 n^3 )= $

    $\dfrac{n (2 n+1) (7 n+1)}{6}$
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