Use the principle of mathematical induction to prove that $ t_n= \frac{18}{3^n(n-1)!}$ for all $n∈ \mathbb Z^+.$

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I am really struggling how to prove that t_n = $\frac{18}{3^n(n-1)!}$ for the sequence where t₁ = 6 and t_n+1 = $\frac{tn}{3n}$

I cannot seem to get past the first step of induction. My method to solve the problem so far includes:

(1) If n = 1, LHS = 6

RHS = t₁

= $\frac{18}{3^1(1-1)!}$

= $\frac{18}{3}$

= 6

LHS = RHS

∴p₁ is true

Assuming p_k is true, n=k

∴ p_k = $\frac{18}{3^k(k-1)!}$

(2) If p_k+1 is true, n=k+1

∴ t_k+1= $\frac{18}{3^k+1(k)!}$

Specifically I am not sure how to show the last line is the case using p_k

Help would be much appreciated! Thank you in advance.

Answer 1

You assumed that $t_k=\frac{18}{3^k(k-1)!}$ is true for all $k\le n$

It’s left to prove that it is true for $n+1$

We have $t_{n+1}=\frac{t_n}{3n}=\frac{\frac{18}{3^n(n-1)!}}{3n}=\frac{18}{3^{n+1}\times n!}$

Since $n(n-1)!=n!$ and $3\times 3^n=3^{n+1}$

And we are done