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| − | The basis case is when <math>n = 0</math><br>
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| − | :<math>\sum_{i=1}^0 i^2 = 0^2 = 0 </math><br>
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| − | and using
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| − | <math>n=0</math>
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| − | in the formula
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| − | <math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
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| − | you get:<br>
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| − | :<math>\frac {0(0 + 1)(2 \cdot 0 + 1)} {6} = \frac {0(1)(0 + 1)} {6} = \frac {0} {6} = 0</math><br>
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| − | Since these are equal, the basis case is true.<br>
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| − | Now, I will show that on the assumption that the summation is true for ''n'', it follows that it is true for <math>n+1</math>
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| − | :<math>\sum_{i = 1}^{n + 1} i^2 =
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| − | (n + 1)^2 + \sum_{i = 1}^n i^2 =
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| − | (n^2 + 2n + 1) + \frac{n(n + 1)(2 \cdot n + 1)} {6} =
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| − | \frac{6n^2 + 12n + 6}{6}+ \frac{2n^3 + 3n^2 + n} {6} =
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| − | \frac {2n^3 + 9n^2 + 13n + 6} {6}</math><br>
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| − | Which should be equal to the formula
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| − | <math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
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| − | when <math>n = n + 1</math>:
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| − | <math>\frac {(n + 1)(n + 1 + 1)(2 \cdot (n + 1) + 1)} {6}=
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| − | \frac {(n + 1)(n + 2)(2n + 2 + 1)} {6}=
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| − | \frac {(n^2 + 3n + 2)(2n + 3)} {6}=
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| − | \frac {2n^3 + 9n^2 + 13n + 6} {6}</math>
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| − | [[introduction-TADM2E|Back to ''Introduction ...'']]
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