A question on sequence.

Question

Given a sequence $$u_n = \frac{\Bigl(1+\bigl(-1\bigr)^n\Bigr)+1}{5n+6}$$ Find the number of terms of the sequence $u_n$ which satisfy the condition $ u_n \in \Bigl(\frac{1}{100},\frac{39}{100}\Bigr)$

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My approach :

First I considered n to a odd natural number . I got the following inequality

$$\frac{1}{100} < \frac{1}{5n+6} < \frac{39}{100} $$ Then considering n to be a even natural number : $$ \frac{1}{100} < \frac{3}{5n+6} < \frac{39}{100}$$

In both the cases I approximated the values of $n$, then found the number of odd and even natural number that lies within that range. My answer was 38. But the correct answer is 18. So please mention those range of odd $n$ values and even $n$ values , it will be very helpful for me and i can find where i did the mistake. And if there is a different approach to this question then please mention that too. I will be glad to know.

Answer 1

HINT: Consider odd $n$ and solve simple inequalities like: $$\frac 1 {100} < \frac 1 {5n+ 6} < \frac {39} {100}.$$

Do the same for even $n$.

Answer 2

Hint: For $n=2m$ we get

$$\frac{1}{100}<\frac{3}{10m+6}<\frac{39}{100}$$ this becomes $$m<29.4$$ Can you proceed now?