# Find the sum of the following series up to 9th term:

2√3+9÷√3+12÷√3+....

## Is this a continued fraction? If not, do you mean what you typed?

2√3 + 9÷√3 + 12÷√3 ...

= √3 (2+3+...+10)

I'm sure you can add up an arithmetic sequence, right?

## To find the sum of the series up to the 9th term, we need to determine the pattern of the terms.

The series can be written as:

2√3 + 9/√3 + 12/√3 + ...

If we simplify each term, we have:

2√3 = (2*√3*√3)/(√3) = (2*3)/(√3) = 6/√3

9/√3 = (9√3)/(√3) = 9√3/√3 = 9

12/√3 = (12√3)/(√3) = 12√3/√3 = 12

We can observe that the terms alternate between 6/√3 and 9, starting with 6/√3.

Now, let's write out the first 9 terms to verify this pattern:

6/√3, 9, 12, 6/√3, 9, 12, 6/√3, 9, 12

We can group the terms into sets of three:

(6/√3 + 9 + 12) + (6/√3 + 9 + 12) + (6/√3 + 9 + 12)

We can simplify each set to find their sum:

(6/√3 + 9 + 12) = 6/√3 + 21

Therefore, the sum of each set of three terms is 6/√3 + 21.

To find the sum of the series up to the 9th term, we need to find how many sets there are. Since each set contains three terms, we divide the total number of terms (9) by 3:

9 terms / 3 terms per set = 3 sets

Now, we can multiply the sum of each set by the number of sets:

3 sets * (6/√3 + 21) = 18/√3 + 63

Thus, the sum of the series up to the 9th term is 18/√3 + 63.

## To find the sum of the series up to the 9th term, we need to determine the pattern and then apply the formula for the sum of an arithmetic series.

Let's examine the given series:

2√3, 9÷√3, 12÷√3, ...

If we simplify the terms, we have:

2√3 = (2/1)√3

9÷√3 = (9/1)÷√3

12÷√3 = (12/1)÷√3

From this, we can observe that the numerators of the terms form an arithmetic sequence: 2, 9, 12, ...

The common difference is 7 (9 - 2 = 7).

Now, let's compute the 9th term of this arithmetic sequence:

T9 = a + (n - 1)d

T9 = 2 + (9 - 1)7

T9 = 2 + 8 * 7

T9 = 2 + 56

T9 = 58

To find the sum of the series up to the 9th term, we use the formula for the sum of an arithmetic series:

Sn = (n/2)(a + Tn)

Sn = (9/2)(2 + 58)

Sn = (9/2)(60)

Sn = 9 * 30

Sn = 270

Therefore, the sum of the given series up to the 9th term is 270.