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!
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.