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giovkast

  • 2 years ago

solve with full induction: sum_{k=1}^{n} k^2 = 1/6n(n + 1)(2n+1)

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  1. giovkast
    • 2 years ago
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    \[\sum_{k=1}^{n} k^2 = 1/6n(n + 1)(2n+1)\]

  2. sirm3d
    • 2 years ago
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    first step: n = 1: \[\sum_{k=1}^{1}k^2=1^2 = 1,\quad (1/6)(1)(2)(3) = 1\]

  3. sirm3d
    • 2 years ago
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    step 2: induction process. assume true for n=s \[\sum_{k=1}^{s}=1^2+2^2+\cdots +s^2=(1/6) s(s+1)(2s+1)\] show the formula is true for n = s+1 \[\sum_{k=1}^{s+1}=1^2+2^2+\cdots + s^2 + (s+1)^2\]\[=\left[ (1/6)s(s+1)(2s+1) \right]+(s+1)^2\]

  4. giovkast
    • 2 years ago
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    here i got stuck

  5. sirm3d
    • 2 years ago
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    simplify the expressions, then factor. if you get \[(1/6)(s+1)(s+2)(2s+3)\] then you're in the right track.

  6. giovkast
    • 2 years ago
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    \[1/3n^3+3/2n2+3/2n+1\]

  7. giovkast
    • 2 years ago
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    i got this, don't know where i went wrong

  8. Jonask
    • 2 years ago
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    you need to add \[\frac{s(s+1)(2s+1)}{6}+\frac{(s+1)^2}{1}=\frac{s(s+1)(2s+1)+6(s+1)^2}{6}\] \[=\frac{(s+1)[s(2s+1)+6(s+1)]}{6}=\frac{(s+1)(2s^2+7s+6)}{6}=\frac{(s+1)(2s+3)(s+2)}{6}\] we need to show that this is the same as n=s+1 in the original ie \[\frac{(\color{red}{s+1})(2\color{red}{(s+1)}+1)(\color{red}{s+1}+1)}{6}\]TRUE

  9. giovkast
    • 2 years ago
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    thank you very much

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