Find a formula for the nth term of the sequence where a_n is calculated directly from n. 3 / 1, 5...

We are given the sequence:

{eq}\displaystyle \dfrac 3 1, \ \dfrac 5 2, \ \dfrac 7 6,\ \dfrac 9{24},\ \dfrac {11}{120},\ \cdots {/eq}

We can approach this problem by looking at the numerators and denominators separately. Note that the first few terms of the numerator are the first few odd numbers starting from 3:

{eq}\displaystyle 3,\ 5,\ 7,\ 9,\ 11,\ ... {/eq}

To generalize this term starting from {eq}\displaystyle n = 1 {/eq}, we can maybe write:

{eq}\displaystyle 2n + 1 {/eq}

We can verify that the first few values of {eq}\displaystyle n {/eq} yield our numerator values:

• {eq}\displaystyle n = 1 \to {/eq} {eq}\displaystyle 2n+1 = 3 {/eq}

• {eq}\displaystyle n = 2 \to {/eq} {eq}\displaystyle 2n+1 = 5 {/eq}

• {eq}\displaystyle n = 3 \to {/eq}{eq}\displaystyle 2n+1 = 7 {/eq}

As for the denominator, the first few terms actually line up with factorials:

• {eq}\displaystyle 1! = 1 {/eq}

• {eq}\displaystyle 2! = 2 {/eq}

• {eq}\displaystyle 3! = 6 {/eq}

• {eq}\displaystyle 4! = 24 {/eq}

• {eq}\displaystyle 5! = 120 {/eq}

So, we can generally write the denominator as:

{eq}\displaystyle n! {/eq}

Therefore, we combine our numerator and denominator sequence term to get:

{eq}\displaystyle \boxed{a_n = \frac{2n+1}{n!}} {/eq}

This works for any positive value of {eq}\displaystyle n {/eq} starting from {eq}\displaystyle n = 1 {/eq}.