Sum of the series `1 + 2 + 4 + 7 +…+ 67` is equal to

Options

A. 150
B. 230
C. 298
D. 340

Answer

To find the sum of the series \(1 + 2 + 4 + 7 + \ldots + 67\), we first need to identify the pattern in the series.

### Step 1: Identify the pattern in the series
The series starts with:
- \(1\) (1st term)
- \(2\) (2nd term, which is \(1 + 1\))
- \(4\) (3rd term, which is \(2 + 2\))
- \(7\) (4th term, which is \(4 + 3\))
- Continuing this, we can see that the differences between consecutive terms are increasing by 1.

The differences are:
- \(2 - 1 = 1\)
- \(4 - 2 = 2\)
- \(7 - 4 = 3\)

This suggests that the \(n\)-th term can be expressed as:
\[ a_n = a_{n-1} + (n-1) \]
where \(a_1 = 1\).

### Step 2: Write the general formula for the \(n\)-th term
From the pattern, we can derive the formula for the \(n\)-th term:
\[ a_n = 1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2} \]
This simplifies to:
\[ a_n = 1 + \frac{n^2 - n}{2} = \frac{n^2 + n + 2}{2} \]

### Step 3: Find \(n\) such that \(a_n = 67\)
Now we need to find \(n\) such that:
\[ \frac{n^2 + n + 2}{2} = 67 \]
Multiplying both sides by 2 gives:
\[ n^2 + n + 2 = 134 \]
This simplifies to:
\[ n^2 + n - 132 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = 1\), and \(c = -132\):
\[ n = \frac{-1 \pm \sqrt{1 + 528}}{2} = \frac{-1 \pm \sqrt{529}}{2} = \frac{-1 \pm 23}{2} \]
Calculating the two possible values:
1. \(n = \frac{22}{2} = 11\)
2. \(n = \frac{-24}{2} = -12\) (not applicable since \(n\) must be positive)

Thus, \(n = 11\).

### Step 5: Calculate the sum of the series
To find the sum of the first \(n\) terms, we can use the formula for the sum of the first \(n\) terms of the series:
\[ S_n = \sum_{k=1}^{n} a_k \]
Using the derived formula for \(a_k\):
\[ S_n = \sum_{k=1}^{n} \left(1 + \frac{k^2 + k + 2}{2}\right) \]
This can be simplified to:
\[ S_n = n + \frac{1}{2} \sum_{k=1}^{n} (k^2 + k + 2) \]
Using the formulas for the sums:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
Substituting \(n = 11\):
\[ S_{11} = 11 + \frac{1}{2} \left( \frac{11 \cdot 12 \cdot 23}{6} + \frac{11 \cdot 12}{2} + 2 \cdot 11 \right) \]
Calculating each part:
- \(\sum_{k=1}^{11} k^2 = \frac{11 \cdot 12 \cdot 23}{6} = 506\)
- \(\sum_{k=1}^{11} k = \frac{11 \cdot 12}{2} = 66\)
- \(2 \cdot 11 = 22\)

Thus:
\[ S_{11} = 11 + \frac{1}{2} \left( 506 + 66 + 22 \right) = 11 + \frac{594}{2} = 11 + 297 = 308 \]

### Final Answer
The sum of the series \(1 + 2 + 4 + 7 + \ldots + 67\) is \(308\).

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