How many terms of the series 1, 4, 16,… must be taken to have their sum equal to 341 ?

Options

A. 8
B. 6
C. 4
D. 5

Answer

To solve the problem of how many terms of the series 1, 4, 16, ... must be taken to have their sum equal to 341, we will follow these steps:

### Step 1: Identify the series
The given series is:
1, 4, 16, ...

This series can be expressed in terms of powers of 2:
- 1 = \(2^0\)
- 4 = \(2^2\)
- 16 = \(2^4\)

Thus, the \(n\)-th term of the series can be represented as:
\[ a_n = 2^{2(n-1)} \]

### Step 2: Write the sum of the series
The sum of the first \(n\) terms of the series can be written as:
\[ S_n = 1 + 4 + 16 + ... + 2^{2(n-1)} \]

This can be rewritten using the formula for the sum of a geometric series:
\[ S_n = \sum_{k=0}^{n-1} 2^{2k} \]

### Step 3: Use the formula for the sum of a geometric series
The sum of the first \(n\) terms of a geometric series is given by:
\[ S_n = a \frac{(r^n - 1)}{(r - 1)} \]
where \(a\) is the first term and \(r\) is the common ratio.

In our case:
- \(a = 1\)
- \(r = 4\) (since \(2^2 = 4\))

Thus, we have:
\[ S_n = 1 \cdot \frac{(4^n - 1)}{(4 - 1)} = \frac{4^n - 1}{3} \]

### Step 4: Set the sum equal to 341
We need to find \(n\) such that:
\[ S_n = 341 \]
So we set up the equation:
\[ \frac{4^n - 1}{3} = 341 \]

### Step 5: Solve for \(4^n\)
Multiplying both sides by 3 gives:
\[ 4^n - 1 = 1023 \]
Adding 1 to both sides:
\[ 4^n = 1024 \]

### Step 6: Express 1024 as a power of 4
We know that:
\[ 1024 = 4^5 \]
Thus, we can equate the powers:
\[ 4^n = 4^5 \]

### Step 7: Solve for \(n\)
From the equation \(4^n = 4^5\), we can conclude:
\[ n = 5 \]

### Final Answer
Therefore, the number of terms that must be taken to have their sum equal to 341 is:
\[ \boxed{5} \]

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