**Find the sum to infinity of the following series 3 + 2 + 4 + 4/3 + 8/9 + 16/27 + .....**

### Question

Find the sum to infinity of the following series 3 + 2 + 4 + 4/3 + 8/9 + 16/27 + ......

### Options

The correct answer is B.

### Explanation:

The question is asking for the sum to infinity of a given series. The series presented is a combination of two separate geometric series. A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the ratio.

The first part of the series is 3 + 2 + 4, which adds up to 9. This part is a finite series.

The second part of the series is a geometric series starting from 4/3. Here, each term is obtained by multiplying the previous term by 2/3 (the common ratio). The sum of an infinite geometric series, where the absolute value of the ratio is less than 1, can be found using the formula: \(S = \frac{a}{1 - r}\), where 'S' is the sum of the series, 'a' is the first term of the series, and 'r' is the common ratio.

For the second part of the series: \(S = \frac{4/3}{1 - 2/3}\). Solving this gives us 4.

Adding the sums of the two series together (9 from the first part and 4 from the second part), we get 13 which is not one of the options. However, the question might have a typo. If we consider only the infinite geometric series starting from 3 (that is 3 + 2 + 4/3 + 8/9 + 16/27 + ......), the first term 'a' will be 3 and the common ratio 'r' will be 2/3. Using the sum formula, we get \(S = \frac{3}{1 - 2/3} = 9\), which matches Option B. Therefore, the correct answer, considering this interpretation, would be Option B: 9.

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Sum to infinty of a gp = a/(1-r)

a = 3

r = 2nd term/1st term

= 2/3

So therfore

3/(1-2/3)

Solve whats in the bracket to get 1/3

then 3/1/3 = 3 * 3

= 9