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

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!

• 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. – fgp May 3 '14 at 23:12
• @fgp Ah, yes. That's the part I thought I messed up. Wouldn't it be: $k(6k+5)+12k+11$ ? – Learner May 3 '14 at 23:23
• $S_{k+1}=S_{k}+12(k+1)-1=k(6k+5)+12(k+1)-1=6k^2+12k+12+5k-1=6(k^2+2k+1)+5(k+1)=6(k+1)^2+5(k+1)$ – user137481 May 3 '14 at 23:37
• 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. – drawnonward May 3 '14 at 23:37

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.
• @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! – fgp May 4 '14 at 0:10
• @user125736 $k(6k+5)$ seems sensible... But again, this is an absolutely gruesome way to teach induction proofs. I'm appalled! – fgp May 4 '14 at 0:30