How many terms of the series `2+6+18+` ... Must be taken to make the sum equal to 728 ?

Answer

To solve the problem of how many terms of the series \(2 + 6 + 18 + \ldots\) must be taken to make the sum equal to 728, we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \(a\) of the series is 2. The second term is 6, and the third term is 18. We can find the common ratio \(r\) by dividing the second term by the first term:
\[ r = \frac{6}{2} = 3 \]

### Step 2: Write the formula for the sum of the first \(n\) terms of a geometric series
The formula for the sum \(S_n\) of the first \(n\) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Substituting the values of \(a\) and \(r\):
\[ S_n = 2 \frac{3^n - 1}{3 - 1} = 2 \frac{3^n - 1}{2} = 3^n - 1 \]

### Step 3: Set up the equation for the sum
We need the sum \(S_n\) to equal 728:
\[ 3^n - 1 = 728 \]

### Step 4: Solve for \(3^n\)
Adding 1 to both sides gives:
\[ 3^n = 728 + 1 = 729 \]

### Step 5: Express 729 as a power of 3
We know that:
\[ 729 = 3^6 \]
Thus, we can equate the powers:
\[ 3^n = 3^6 \]

### Step 6: Solve for \(n\)
Since the bases are the same, we can set the exponents equal to each other:
\[ n = 6 \]

### Conclusion
Therefore, the number of terms that must be taken to make the sum equal to 728 is \(n = 6\).

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