Mathematical induction - recursion

Question

I am struggling with that exercise... i don't understand a step in the answer. Maybe somebody Can explain in here or at least give me a hint what's going on there.

$a_{n} = \frac{7*(n+2)}{2*n} *a_{n-1}$ ,$a_{1} = 14$

So it's...

$a_{2} = \frac{7*4}{4} * a_{1} = \frac{7*4 * 14}{4} = \frac{2* 7^{2}*4 }{4}$

$a_{3} = \frac{7*5}{6} * a_{2} = \frac{2* 7^{3}*4*5 }{4*6}$

And here is the problem... a4 equals:

$a_{4} = \frac{7*6}{8} * a_{3} = \frac{2* 7^{4}*4*5*6 }{4*6*8}$

We are multiplying numerator and denominator with 2*3 to get factorial 6! on numerator. I can't understand how the $(1*2*3)^2$ got on denominator and how after that we transformed it to $3! * 4!$. Next steps are quite easy and the mathematical induction is fully understandable for me.

$a_{4} = \frac{2* 7^{4}*6! }{ (1*2*3)^{2} * 2 *3 *4 } = \frac{2* 7^{4}*6! }{ 3! 4! }$

Full answer of the exercise: JPG image

Answer 1

Your doubt is valid: $$\begin{align}a_{n}& = \frac{7*(n+2)}{2*n} *a_{n-1}, a_1=14\\ a_2&=\frac{7*4}{4}*a_1=\frac{2*7^2*4}{4}=\frac{2*7^2*4!}{4!}\\ a_3&=\frac{7*5}{6}*a_2=\frac{2*7^3*4*5}{4*6}=\frac{7^3*5!}{2*3!*3!}\\ a_4&=\frac{7*6}{8}*a_3=\frac{7^4*5!*6}{2*3!*3!*8}=\frac{7^4*6!}{2^2*3!*4!}\\ a_5&=\frac{7*7}{10}*a_4=\frac{7^5*6!*7}{2^2*3!*4!*10}=\frac{7^5*7!}{2^3*3!*5!}\\ \vdots\\ a_n&=\frac{7^n*(n+2)!}{2^{n-2}*3!*n!},n\ge 1.\end{align}$$ Verify: $$\begin{align}a_1&=\frac{7^1*(1+2)!}{2^{1-2}*3!*1!}=14\\ a_2&=\frac{7^2*(2+2)!}{2^{2-2}*3!*2!}=2*7^2\\ a_3&=\frac{7^3*(3+2)!}{2^{3-2}*3!*3!}=\frac{7^3*5!}{2*3!*3!}\end{align}$$

Answer 2

$$a_n=a_{n-r}\cdot\dfrac{7^r(n+2)(n+1)\cdots(n-r+3)}{2^r n(n-1)\cdots(n-r+1)}$$ $0\le r\le n-1$

For $r\ge2\implies n\ge3$ $$a_n=a_{n-r}\cdot\left(\dfrac72\right)^r\dfrac{(n+2)(n+1)}{(n-r+2)(n-r+1)}$$

Set $r=n-1$

Answer 3

I got $$a_n=\frac{2}{3}\frac{7^n}{2^n}(n+1)(n+2)$$