Find the pattern of a series

Question

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.

Greg V · Accepted Answer · 2021-03-15 02:20:34Z

2

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

answered Mar 15, 2021 at 2:20

Greg V

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2

$\begingroup$ (+1) This is extremely useful for applying the ratio test. $\endgroup$
– robjohn ♦
Mar 15, 2021 at 2:23
$\begingroup$ 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 $\endgroup$
– Sirou Ewei
Mar 15, 2021 at 2:29
1

$\begingroup$ 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 $\endgroup$
– Greg V
Mar 15, 2021 at 2:30
$\begingroup$ 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? $\endgroup$
– Sirou Ewei
Mar 15, 2021 at 3:10
$\begingroup$ That looks spot on, except maybe an index or two being off. What's the next step of the ratio test? $\endgroup$
– Greg V
Mar 15, 2021 at 3:11

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