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| − | Call the statement <math>S_n</math> and the general term <math>a_n</math><br>
| + | #REDIRECT [[UNTV]] |
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| − | <b>Step 1:</b> Show that the statement holds for the basis case <math>n = 0</math><br>
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| − | :<math>a_0 = 0 \cdot (0 + 1)(0 + 2) = 0</math>
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| − | <br><br>
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| − | :<math>S_0 = \frac {0 \cdot (0 + 1)(0 + 2)(0 + 3)} {4} = \frac {0} {4} = 0 </math><br>
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| − | Since <math>a_n = S_n</math>, the basis case is true.<br><br>
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| − | <b>Step 2:</b> Assume that <math>n = k</math> holds.
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| − | :<math>S_k = \frac {k(k + 1)(k + 2)(k + 3)} {4}</math><br><br>
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| − | <b>Step 3:</b> Show that on the assumption that the summation is true for ''k'', it follows that it is true for ''k + 1''.
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| − | :<math>
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| − | \begin{align}
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| − | S_{k + 1} & = \sum_{i = 1}^k i(i + 1)(i + 2) + a_{k + 1} \\
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| − | & = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)((k + 1) + 1)((k + 1) + 2) \\
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| − | & = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)(k + 2)(k + 3) \\
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| − | & = \frac{k(k + 1)(k + 2)(k + 3)}{4} + \frac{4 \cdot (k + 1)(k + 2)(k + 3)} {4} \\
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| − | \end{align}
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| − | </math><br>
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| − | It's easier to factor than expand. Notice the common factor of ''(k + 1)(k + 2)(k + 3)''.
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| − | :<math>S_{k + 1} = \frac{(k + 1)(k + 2)(k + 3)(k + 4)}{4}</math><br>
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| − | This should be equal to the formula
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| − | <math>S_k = \frac{k(k + 1)(k + 2)(k + 3)}{4}</math>
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| − | when ''k = k + 1'':<br>
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| − | :<math>\frac{(k + 1)((k + 1) + 1)((k + 2) + 2)((k + 1) + 3)} {4} =
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| − | \frac{(k + 1)(k + 2)(k + 3)(k + 4)} {4}</math><br><br>
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| − | Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that S_n holds for all natural n
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| − | [[introduction-TADM2E|Back to ''Introduction ...'']] | |
