# Find the nth term of 1/3+4/243+1/243+16/19,683?

• GiannaMe'arah
In summary, the given sequence follows a pattern where the numerators increase by powers of 4 and the denominators increase by powers of 81. To find the nth term, we can use the formula a_n = a_1 * r^(n-1), where a_n is the nth term, a_1 is the first term, and r is the common ratio. The common ratio in this sequence is 4/81. To simplify the sequence, we can rewrite the fractions as powers of 3 and 81, combine like terms, and use the rule a^-b = 1/a^b. The general formula for the nth term in this type of sequence is a_n = a_1 * (x/y
GiannaMe'arah
by simplifying the denominator, 1/3 + 4/3^5 + 9/3^7 + 16/3^9

Cant find the nth term. i am assuming there is something wrong with the given. i suppose 1/3 should be 1/3^3. If so, then the nth term is n^2/3^(2n+1)

thanks.. :)

In your title you wrote 1/3+4/243+1/243+16/19,683?

Was the third term a typo? If it's how you write it in your post, then yeah, I think the first should be 1/35 instead.

## 1. What is the pattern in the given sequence?

The pattern in the given sequence is that the numerators are increasing by powers of 4, while the denominators are increasing by powers of 81.

## 2. How can I find the nth term in this sequence?

To find the nth term in this sequence, we can use the formula an = a1 * r^(n-1), where an is the nth term, a1 is the first term, and r is the common ratio. In this sequence, a1 = 1/3 and r = 4/81.

## 3. What is the common ratio in this sequence?

The common ratio in this sequence is 4/81.

## 4. How can I simplify the given sequence?

To simplify the given sequence, we can first rewrite the fractions as powers of 3 and 81. This gives us 1/3 = 3^-1 and 4/81 = 3^-4. Then, we can combine the like terms to get 3^-1 + 3^-4 + 3^-2 + 3^-8. Finally, using the rule a^-b = 1/a^b, we can simplify the sequence to 3^-1 + 3^-4 + 3^-2 + 3^-8 = 1/3 + 1/81 + 1/9 + 1/59049.

## 5. What is the general formula for the nth term in this type of sequence?

The general formula for the nth term in a sequence where the numerators increase by powers of x and the denominators increase by powers of y is an = a1 * (x/y)^(n-1). In the given sequence, x = 4 and y = 81, so the formula becomes an = a1 * (4/81)^(n-1).

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