Consider sequence $ d _ 1, d _2, d_3 $
$$d_k= \frac {d_{k-1}} {k + 1} $$ for all integers $k \ge 2 $ with the initial condition that $ d_1 = 1$.
Find an explicit formula for $d_k$ for the $k^{th}$ term.
So far I have figured out $d_1 = 1, d_2 = \frac {1}{3}, d_3 = \frac {1}{12}$
I am not sure how to solve for the explicit formula. If someone could explain a little further.
On
Note that at each progressive stage, you keep multiplying by $\frac1n$ for successive $n$. So $d_2=\frac{1}{3},$ $d_3=\frac{1}{4\cdot 3},$ and in general for larger $n$ $$d_{n}=\frac{1}{(n+1)\cdot n\cdots 4\cdot 3}=\frac{2}{(n+1)\cdot n\cdots 4\cdot 3\cdot 2\cdot 1}=\frac{2}{(n+1)!}.$$
You'll want to show that this holds for all $n\ge1.$ The proof by induction is straightforward.
To find an explicit formula, you may calculate a few more terms and then you use all your maths experience in the past to match this to $$d_k = \frac{2}{(k+1)!}$$
You can then proceed to prove that this formula is true by using induction. Remember any time you keep multiplying a number by one higher, a factorial will be involved.