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<h1 align="center">Mathematical Induction</h1>
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<div class="center80">
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<p>Mathematical Induction is a special way of proving things. It has only 2 steps:</p>
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<ul>
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<li>Step 1. Show it is true for the <b>first one</b></li>
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<li>Step 2. Show that if <b>any one</b> is true then the <b>next one</b> is true</li>
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</ul>
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<p>Then <b>all</b> are true</p>
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</div><p> </p>
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<p style="float:left; margin: 0 30px 5px 0;"><img src="images/domino-effect.jpg" alt="Domino Effect" width="250" height="158" /></p>
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<p class="larger">Have you heard of the "Domino Effect"?</p>
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<ul>
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<li>Step 1. The <b>first</b> domino falls</li>
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<li>Step 2. When <b>any</b> domino falls, the <b>next</b> domino falls</li>
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</ul>
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<p>So ... <b>all dominos will fall!</b></p>
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<p class="larger">That is how Mathematical Induction works.</p>
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<p>In the world of numbers we say:</p>
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<ul>
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<li>Step 1. Show it is true for first case, usually <b>n=1</b></li>
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<li>Step 2. Show that if <b>n=k</b> is true then <b>n=k+1</b> is also true</li>
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</ul>
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<h2>How to Do it</h2>
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<p>Step 1 is usually easy, we just have to prove it is true for <b>n=1</b></p>
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<p>Step 2 is best done this way:</p>
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<ul><li><b>Assume</b> it is true for <b>n=k</b></li>
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<li><b>Prove</b> it is true for <b>n=k+1</b> (we can use the <b>n=k</b> case as a <b>fact</b>.)</li>
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</ul>
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<p>It is like saying <i><b>"IF we can make a domino fall, WILL the next one fall?"</b></i><b></b></p>
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<p> </p>
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<p>Step 2 can often be <i><b>tricky</b></i>, we may need to use imaginative tricks to make it work! </p>
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<p>Like in this example:</p>
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<div class="example">
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<h3>Example: is 3<sup>n</sup>−1 a multiple of 2?</h3>
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<p><b>Is that true? Let us find out.</b></p>
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<p> </p>
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<p class="larger"><b>1.</b> Show it is true for <b>n=1</b></p>
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<p align="center"><b>3<sup>1</sup>−1 = 3−1 = 2 </b></p>
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<p>Yes 2 is a multiple of 2. That was easy.</p>
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<p align="center" class="large">3<sup>1</sup>−1 is true</p>
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<p class="larger"> </p>
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<p class="larger"><b>2.</b> Assume it is true for <b>n=k</b></p>
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<p align="center" class="large">3<sup>k</sup>−1 is true </p>
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<p align="center">(Hang on! How do we know that?
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We don't! <br />
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It is an <b>assumption</b> ... that we treat <br />
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<b>as a fact</b> for the rest of this example)</p>
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<p class="larger"> </p>
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<p class="larger">Now, prove that <b>3<sup>k+1</sup>−1</b> is a multiple of 2</p>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/mathematical-induction.svg" alt="mathematical induction a" /></p><p> </p>
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<p><b>3<sup>k+1</sup></b> is also <b>3×3<sup>k</sup></b></p>
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<p>And then split <b>3×</b> into <b>2×</b> and <b>1×</b></p>
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<p>And each of these are multiples of 2</p>
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<p class="larger"> </p>
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<div class="example2"></div>
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<p>Because:</p>
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<ul>
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<li><b>2×3<sup>k</sup></b> is a multiple of 2 (we are multiplying by 2)</li>
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<li><b>3<sup>k</sup>−1 is true </b>(we said that in the assumption above)</li>
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</ul>
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<p>So:</p>
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<p align="center"><span class="large">3<sup>k+1</sup>−1 is true </span></p>
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<p>DONE!</p>
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</div>
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<p>Did you see how we used the <b>3<sup>k</sup>−1</b> case as being <b>true</b>, even though we had not proved it? That is OK, because we are relying on the <b>Domino Effect</b> ... </p>
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<p align="center">... we are asking <b>if</b> any domino falls will the <b>next one</b> fall?</p>
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<p>So we take it as a fact (temporarily) that the "<b>n=k</b>" domino falls (i.e. <b>3<sup>k</sup>−1</b> is true), and see if that means the "<b>n=k+1</b>" domino will also fall.</p>
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<h2>Tricks</h2>
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<p>I said before that we often need to use imaginative tricks. </p>
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<div class="center80">
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<p>A common trick is to rewrite the <b>n=k+1</b> case into 2 parts:</p>
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<ul>
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<li>one part being the <b>n=k</b> case (which is assumed to be true)</li>
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<li>the other part can then be checked to see if it is also true</li>
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</ul>
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</div>
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<p>We did that in the example above, and here is another one:</p>
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<div class="example">
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<h3>Example: Adding up Odd Numbers </h3>
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<p align="center" class="larger">1 + 3 + 5 + ... + (2n−1) = n<sup>2</sup></p>
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<p class="larger"><b>1.</b> Show it is true for <b>n=1</b></p>
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<p class="center large">1 = 1<sup>2</sup> is True</p>
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<p class="larger"> </p>
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<p class="larger"><b>2.</b> Assume it is true for <b>n=k</b></p>
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<p align="center" class="large"></p>
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<p align="center"> <span class="large">1 + 3 + 5 + ... + (2k−1) = k<sup>2</sup> is True</span><br>
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(An assumption!)</p>
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<p class="larger">Now, prove it is true for "k+1"</p>
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<p align="center" class="larger">1 + 3 + 5 + ... + (2k−1) + (2(k+1)−1) = (k+1)<sup>2</sup> <b>?</b></p>
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<p> </p>
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<p>We know that <b>1 + 3 + 5 + ... + (2k−1) = k<sup>2</sup></b> (the assumption above), so we can do a replacement for all but the last term:</p>
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<p align="center" class="larger"><b>k<sup>2</sup></b> + (2(k+1)−1) = (k+1)<sup>2</sup></p>
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<p>Now expand all terms:</p>
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<p align="center" class="larger">k<sup>2</sup> + 2k + 2 − 1 = k<sup>2</sup> + 2k+1 </p>
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<p>And simplify: </p>
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<p align="center" class="larger">k<sup>2</sup> + 2k + 1 = k<sup>2</sup> + 2k + 1 </p>
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<p align="center"><b>They are the same! So it is true.</b></p>
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<p>So:</p>
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<p class="center large">1 + 3 + 5 + ... + (2(k+1)−1) = (k+1)<sup>2</sup> is True</p>
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<p>DONE!</p>
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</div>
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<p> </p>
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<h2>Your Turn</h2>
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<p>Now, here are two more examples <b>for you to practice</b> on. </p>
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<p>Please try them first yourself, then look at our solution below.</p>
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<div class="example">
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<h3>Example: Triangular Numbers</h3><p><a href="triangular-numbers.html">Triangular numbers</a> are numbers that can make a triangular dot pattern.</p><p class="center"><img src="../numbers/images/triangular-number-dots.svg" alt="triangular numbers" style="max-width: 100%;" /></p>
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<p class="center large">Prove that the <b>n-th</b> triangular number is:</p>
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<p class="center large"> T<sub>n</sub> = n(n+1)/2</p>
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</div>
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<div class="example">
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<p style="float:right; margin: 0 0 5px 10px;"><img src="../numbers/images/cube.gif" alt="cube 3x3x3" width="175" height="189" /></p>
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<h3>Example: Adding up Cube Numbers</h3><p><a href="../numbers/cube-root.html">Cube numbers</a> are the cubes of the Natural Numbers</p>
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<p class="center large">Prove that: </p>
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<p class="center large">1<sup>3</sup> + 2<sup>3</sup> + 3<sup>3</sup> + ... + n<sup>3</sup> = ¼n<sup>2</sup>(n + 1)<sup>2</sup></p>
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</div>
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<p class="center"> </p>
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<p class="center">. . . . . . . . . . . . . . . . . .</p>
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<p class="center"> </p>
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<p> </p>
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<p>Please don't read the solutions until you have tried the questions yourself, these are the only questions on this page for you to practice on!</p>
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<div class="example">
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<h3>Example: Triangular Numbers</h3>
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<p class="center"><img src="../numbers/images/triangular-number-dots.svg" alt="triangular numbers" width="80%" /></p>
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<p class="large">Prove that the <b>n-th</b> triangular number is:</p>
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<p class="center large"> T<sub>n</sub> = n(n+1)/2</p>
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<p> </p>
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<p class="larger"><b>1.</b> Show it is true for <b>n=1</b></p>
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<p class="center large">T<sub>1</sub> = 1 × (1+1) / 2 = 1 is True</p>
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<p> </p>
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<p class="larger"><b>2.</b> Assume it is true for <b>n=k</b></p>
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<p class="center">T<sub>k</sub> = k(k+1)/2 is True (An assumption!)</p>
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<p>Now, prove it is true for "k+1"</p>
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<p class="center">T<sub>k+1</sub> = (k+1)(k+2)/2 ?</p>
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<p> </p>
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<p>We know that T<sub>k</sub> = k(k+1)/2 (the assumption above)</p>
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<p class="center">T<sub>k+1</sub> has an extra row of (k + 1) dots</p>
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<p>So, T<sub>k+1</sub> = T<sub>k</sub> + (k + 1)</p>
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<p class="so">(k+1)(k+2)/2 = k(k+1) / 2 + (k+1)</p>
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<p>Multiply all terms by 2:</p>
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<p class="so">(k + 1)(k + 2) = k(k + 1) + 2(k + 1)</p>
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<p class="so">(k + 1)(k + 2) = (k + 2)(k + 1)</p>
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<p>They are the same! So it is <b>true</b>.</p>
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<p>So:</p>
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<p class="center large">T<sub>k+1</sub> = (k+1)(k+2)/2 is True</p>
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<p>DONE!</p>
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</div>
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<p> </p>
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<div class="example">
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<p style="float:right; margin: 0 0 5px 10px;"><img src="../numbers/images/cube.gif" alt="cube 3x3x3" width="175" height="189" /></p>
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<h3>Example: Adding up Cube Numbers</h3>
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<p class="center large">Prove that: </p>
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<p class="center large">1<sup>3</sup> + 2<sup>3</sup> + 3<sup>3</sup> + ... + n<sup>3</sup> = ¼n<sup>2</sup>(n + 1)<sup>2</sup></p>
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<p> </p>
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<p class="larger"><b>1.</b> Show it is true for <b>n=1</b></p>
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<p class="center large">1<sup>3</sup> = ¼ × 1<sup>2</sup> × 2<sup>2</sup> is True</p>
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<p> </p>
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<p class="larger"><b>2.</b> Assume it is true for <b>n=k</b></p>
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<p class="center">1<sup>3</sup> + 2<sup>3</sup> + 3<sup>3</sup> + ... + k<sup>3</sup> = ¼k<sup>2</sup>(k + 1)<sup>2</sup> is True (An assumption!)</p>
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<p>Now, prove it is true for "k+1"</p>
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<p class="center">1<sup>3</sup> + 2<sup>3</sup> + 3<sup>3</sup> + ... + (k + 1)<sup>3</sup> = ¼(k + 1)<sup>2</sup>(k + 2)<sup>2</sup> ?</p>
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<p> </p>
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<p>We know that 1<sup>3</sup> + 2<sup>3</sup> + 3<sup>3</sup> + ... + k<sup>3</sup> = ¼k<sup>2</sup>(k + 1)<sup>2</sup> (the assumption above), so we can do a replacement for all but the last term:</p>
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<p class="center">¼k<sup>2</sup>(k + 1)<sup>2</sup> + (k + 1)<sup>3</sup> = ¼(k + 1)<sup>2</sup>(k + 2)<sup>2</sup></p>
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<p>Multiply all terms by 4:</p>
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<p class="center">k<sup>2</sup>(k + 1)<sup>2</sup> + 4(k + 1)<sup>3</sup> = (k + 1)<sup>2</sup>(k + 2)<sup>2</sup></p>
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<p>All terms have a common factor (k + 1)<sup>2</sup>, so it can be canceled:</p>
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<p class="center">k<sup>2</sup> + 4(k + 1) = (k + 2)<sup>2</sup></p>
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<p>And simplify:</p>
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<p class="center">k<sup>2</sup> + 4k + 4 = k<sup>2</sup> + 4k + 4</p>
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<p>They are the same! So it is true.</p>
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<p>So:</p>
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<p class="center large">1<sup>3</sup> + 2<sup>3</sup> + 3<sup>3</sup> + ... + (k + 1)<sup>3</sup> = ¼(k + 1)<sup>2</sup>(k + 2)<sup>2</sup> is True</p>
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<p>DONE!</p>
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</div>
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<p> </p>
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<div class="related"><a href="index.html">Algebra Index</a></div>
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