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So I'm learning mathematical induction as a proof technique (teaching myself discrete math as a foundation for a comp sci class I'm going to be taking). My algebra is a little rusty, and I cannot follow how this proof gets to the second step of (1/4)k^2(k+1)^2+(k+1)^3. I'm also lost on how we get from line 2 to 3, where it goes from (1/4)K^2(k+1)^2 + (K+1)^3 to (1/4)(k+1)^2[k^2+4(k+1)]. Any help is appreciated!


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It just distributes the square $\bigg[ \frac{k(k+1)}{2}\bigg]^2 = \frac{k^2(k+1)^2}{2^2}$. – Robert Cardona Jan 26 '13 at 6:52
In general, $\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$. More generally still, $\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$. – André Nicolas Jan 26 '13 at 6:53
Ah, cool. How does the 1/4 then get factored out? – user56763 Jan 26 '13 at 7:00
$(\frac {1}{2} )^2 = \frac {1}{4}$. – Calvin Lin Jan 26 '13 at 7:00
Bingo, thanks. I forgot that (a/b)c = ac/b – user56763 Jan 26 '13 at 7:17
up vote 1 down vote accepted

I've added one line with some color

Now L.H.S. is $$ \begin{align} 1^3+2^3+\dots+k^3+(k+1)^3 &=\left[\frac{k(k+1)}{2}\right]^2+(k+1)^3\\ &=\frac14k^2(k+1)^2+(k+1)^3\\ &=\frac14\color{#C00000}{k^2}(k+1)^2+\frac14(k+1)^2\color{#00A000}{4(k+1)}\\ &=\frac14(k+1)^2\left[\color{#C00000}{k^2}+\color{#00A000}{4(k+1)}\right]\\ &=\frac14(k+1)^2\left[k^2+4k+4\right]\\ &=\frac14(k+1)^2(k+2)^2\\ &=\left[\frac{(k+1)(k+2)}{2}\right]^2 \end{align} $$ Therefore L.H.S.=R.H.S

i.e. $\displaystyle1^3+2^3+\dots+k^3+(k+1)^3=\left[\frac{(k+1)(k+2)}{2}\right]^2$

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I apologize for sounding like a broken record, but be aware of the fact that about ten percent of the male population is colorblind to a greater or lesser degree. Blue is a fairly safe choice, but two colors, especially red and green, is less so. – Rick Decker Jan 26 '13 at 20:19
@RickDecker: note that the color is simply an aid. The added line is broken down further to the point where the pieces simply need to be moved to the right spot. – robjohn Jan 26 '13 at 20:48
The extra line made all the difference. I understand proof by induction, but the algebra to apply the inductive hypothesis is still tripping me up. – user56763 Jan 29 '13 at 4:20

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