# Find the pattern of a series

I am given the series $$\frac{1}{6} + \frac{1\cdot 8}{6\cdot 10} + \frac{1\cdot 8\cdot 15}{6\cdot 10\cdot 14}$$

I am asked to find a formula for this series $$a_n$$

So I have found that the last fraction multiplied at $$nth$$ term is $$\frac{(1+7(n-1))}{(6+4(n-1))}$$

But I am stuck, unsure how to proceed after this.

Try recursion.

You know that each term (after the first term, of course) is simply the previous term multiplied by a varying constant that relates to $$n$$.

So try something like:

$$a_1 = \frac{1}{6}$$ and $$a_n = a_{n-1} \cdot \frac{1+7(n-1)}{6+4(n-1)}.$$

• (+1) This is extremely useful for applying the ratio test.
– robjohn
Mar 15, 2021 at 2:23
• Can you elaborate a bit more? The question asks me to use ratio test, which is why I am trying to find a formula for $a_n$ . I tried to substitute $a_n = \frac{1+7n}{6+4n} \cdot \frac{1+7(n-1)}{6+4(n-1)}$ and use ratio test to cancel out the $nth$ term. But I did not succeed Mar 15, 2021 at 2:29
• Not to hijack robjohn's comment, but think about what you need to do to apply the ratio test, and think about how the $n+1^{th}$ term relates to the $n^{th}$ term in the recursive definition Mar 15, 2021 at 2:30
• During ratio test, all the previous terms should cancel out, leaving only $\frac{(1+7(n-1))}{(6+4(n-1))}$. But this is incorrect? Mar 15, 2021 at 3:10
• That looks spot on, except maybe an index or two being off. What's the next step of the ratio test? Mar 15, 2021 at 3:11