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Use the principle of mathematical induction to show that the given statement is true for all natural numbers n.

$S_n: 11+23+35+...+(12n-1)=n(6n+5)$

My work:

$S_1:(12*1-1) \overset?= 1(6*1+5)$

$11 = 11$

$S_k:11+23+35+...+(12k-1)=k(6k+5)$ $S_{k+1}: 11+23+35+...+(12k-1)+12(k+1)-1 \overset?= (k+1)[6(k+1)+5]$

$(12k-1)+1+12k+11 \overset?=(k+1)(6k+11)$

I want to verify that all my work up to this point is correct. Especially the last line, which should be expressed in factored form. Thank you!

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    $\begingroup$ The left-hand side of the last line should be $k(6k + 5) + (12k+1)$. Dunno how you arrived at the expression you wrote, but it can't be right, there's a $k^2$ term missing there. $\endgroup$ – fgp May 3 '14 at 23:12
  • $\begingroup$ @fgp Ah, yes. That's the part I thought I messed up. Wouldn't it be: $k(6k+5)+12k+11$ ? $\endgroup$ – Learner May 3 '14 at 23:23
  • $\begingroup$ $S_{k+1}=S_{k}+12(k+1)-1=k(6k+5)+12(k+1)-1=6k^2+12k+12+5k-1=6(k^2+2k+1)+5(k+1)=6(k+1)^2+5(k+1)$ $\endgroup$ – user137481 May 3 '14 at 23:37
  • $\begingroup$ error, the lhs of the last line should be $k(6k+5) + 12(k+1) - 1 = 6k^2 + 5k +12k + 11$ from here it is an easy factorization. $\endgroup$ – drawnonward May 3 '14 at 23:37
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We want to prove that $$ 11 + 23 + 35 + \ldots + (12n - 1) = n(6n + 5) \text{.} $$

We start with the base case $n=1$, i.e. we have to validate $$ 11 \overset?= 1(6\cdot 1 + 5) $$ which is indeed true.

Now we assume that the statement is true for some $n$ (the induction hypothesis), and using that assumption prove that it's also true for $n+1$. This is the induction step. In other words, we have to show that $$ 11 + \ldots + (12n - 1) = n(6n + 5) \Rightarrow \underbrace{11 + \ldots + (12(n+1) - 1)}_{A} = \underbrace{(n+1)(6(n+1) + 5)}_{B} \text{.} $$ We do that by observing that $$\begin{eqnarray} A &=& 11 + \ldots + (12(n+1) - 1) \\ &=& \underbrace{11 + \ldots + (12n - 1)}_{=n(6n + 5) \,(\star)} + (12(n+1) - 1) \\ &=& 6n^2 + 5n + 12n + 12 - 1 \\ &=& 6n^2 + 17n + 11 \end{eqnarray}$$ and that also $$\begin{eqnarray} B &=& (n+1)(6(n+1) + 5) \\ &=& (n+1)(6n + 11)\\ &=& 6n^2 + 11n + 6n + 11 \\ &=& 6n^2 + 17n + 11 \text{.} \end{eqnarray}$$ So $A=B$, which completes the proof of the induction step. $(\star)$ is where we used the induction hypothesis.

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  • $\begingroup$ I think undergrad's solution is a bit more slick personally, but this is equally valid. $\endgroup$ – drawnonward May 3 '14 at 23:52
  • $\begingroup$ @fgp This is what I initially got as well, but how in the world do I fit it into this format? imgur.com/4o00Pwd I truly appreciate the effort. $\endgroup$ – Learner May 3 '14 at 23:55
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    $\begingroup$ @user125736 That left-hand side in that image already uses the assumption as far as I can see. All that's missing is factoring out the $(k+1)$ term. But whoever forces you to do solve math problems but typing seemingly arbitrary stuff into some form must be quite ouf of this mind.... This is horrible! $\endgroup$ – fgp May 4 '14 at 0:10
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    $\begingroup$ @user125736 There just isn't enough context there.... Is that an image of the whole question? And what are the blue boxes? $\endgroup$ – fgp May 4 '14 at 0:24
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    $\begingroup$ @user125736 $k(6k+5)$ seems sensible... But again, this is an absolutely gruesome way to teach induction proofs. I'm appalled! $\endgroup$ – fgp May 4 '14 at 0:30

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