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I'm trying to solve by induction that $1^3 + 2^3 + ... + (n+1)^3 = [(n(n+1))/2]^2$

However, I have a lot of trouble, and it must be said that I don't do a great deal of mathematics.

I keep getting $(k+1)^3 = (k+1)^2$ every time I try to solve it. Please point out where I am going wrong.

$[ (k(k+1))/2]^2 + (k+1)^3 = [ (k(k+1))/2 + (k+1) ]^2$

$(k^2(k+1)^2)/4 + (k+1)^3 = (k^2(k+1)^2)/4 + (k+1)^2$

Please help.

hunter
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stuck
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1 Answers1

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Please point out where I am going wrong.

$[ (k(k+1))/2]^2 + (k+1)^3 = [ (k(k+1))/2 + (k+1) ]^2$

$(k^2(k+1)^2)/4 + (k+1)^3 = (k^2(k+1)^2)/4 + (k+1)^2$

You did this on the RHS: $$\left(\frac{k(k+1)}2+(k+1)\right)^2 = \frac{k^2(k+1)^2}4 + (k+1)^2.$$ This is not correct. You have $(a+b)^2=a^2+2ab+b^2$, not $(a+b)^2=a^2+b^2$. So you should get $$\left(\frac{k(k+1)}2+(k+1)\right)^2 = \frac{k^2(k+1)^2}4 + k(k+1)^2 + (k+1)^2.$$ You may notice that $k(k+1)^2 + (k+1)^2= k(k+1)^2 + 1(k+1)^2 = (k+1)(k+1)^2 = (k+1)^3$. So you get exactly the equality you wanted.

Community
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Martin Sleziak
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  • Thanks for this, it certainly cleared up my misguided thinkings. I will try this exercise again fresh tomorrow and see how I get on with this in mind! – stuck Feb 16 '15 at 15:16
  • Can you elaborate on the bit at the end, I'm not certain I follow correctly. – stuck Feb 16 '15 at 17:28
  • @stuck I used distributivity, i.e. $ac+bc=(a+b)c$. In this case I used it for $a=k$, $b=1$, $c=(k+1)^3$. – Martin Sleziak Feb 16 '15 at 18:04
  • I meant the whole last note, I got the distributivity of the (k=1)(k+1)^2 but I didn't follow the k(k+1)^2 + 1(k+1)^2 becoming (k+1)(k+1)^2 ... I have never been good at maths and I haven't done any in a long while, so I need quite a lot of detail I'm afraid. – stuck Feb 17 '15 at 21:53
  • @stuck $\color{blue}{k}\color{green}{(k+1)^2} + \color{blue}{1}\color{green}{(k+1)^2} = \color{blue}{(k+1)}\color{green}{(k+1)^2}$ – Martin Sleziak Feb 18 '15 at 07:39
  • I believe it, but can you explain how and why... – stuck Feb 18 '15 at 10:18
  • @stuck Notice the use of colors in my last comment. I colored $(k+1)^2$ in green to stress that this is a common factor which I want to "pull out". – Martin Sleziak Feb 18 '15 at 10:39
  • Just two examples where similar things are used: http://math.stackexchange.com/questions/436810/stuck-on-rearranging-of-this-equation and http://math.stackexchange.com/questions/535008/how-can-i-use-the-distributive-property-to-rewrite-an-algebraic-fraction – Martin Sleziak Feb 18 '15 at 10:52
  • I don't follow how (k+1)^2 + (k+1)^2 became (k+1)^2 ... I will read over the links. – stuck Feb 18 '15 at 11:09
  • @stuck It is not that $(k+1)^2+(k+1)^2$ became $(k+1)^2$. It is $\color{blue}k(k+1)^2+(k+1)^2$. I.e., both summands contain $(k+1)^2$. Hence $(k+1)^2$ is common factor, which I am going to "factor out". – Martin Sleziak Feb 18 '15 at 11:13
  • BTW the above links suggest that it is ok to post also questions about simple algebraic manipulations. It is possible that somebody will be able to give you better explanation if you post this a separate question. (It might be useful to add link to this post, so that it is not closed for the lack of context.) Posting a new question would also give you space to explain in more detail what exactly you're having problems with. – Martin Sleziak Feb 18 '15 at 11:15
  • Thanks for the clarification. Not sure why it looked so strange, seems easier to understand when I draw brackets. Either way, I think I follow now and will try to do some more problems of this nature. Hopefully before long I wont be so helpless! Thanks again. – stuck Feb 18 '15 at 11:23