`99^(th)` term of the series `2 + 7 + 14 + 23 + 34 +…`is

Options

A. 9998
B. 9999
C. 10000
D. none of these

Answer

To find the 99th term of the series \(2, 7, 14, 23, 34, \ldots\), we first need to identify the pattern in the series.

### Step 1: Identify the terms and their differences
The terms of the series are:
- \(T_1 = 2\)
- \(T_2 = 7\)
- \(T_3 = 14\)
- \(T_4 = 23\)
- \(T_5 = 34\)

Now, let's find the differences between consecutive terms:
- \(T_2 - T_1 = 7 - 2 = 5\)
- \(T_3 - T_2 = 14 - 7 = 7\)
- \(T_4 - T_3 = 23 - 14 = 9\)
- \(T_5 - T_4 = 34 - 23 = 11\)

The differences are \(5, 7, 9, 11\), which form an arithmetic progression (AP) with a first term of \(5\) and a common difference of \(2\).

### Step 2: General term of the difference sequence
The \(n\)-th term of the difference sequence can be expressed as:
\[ d_n = 5 + (n - 1) \cdot 2 = 2n + 3 \]

### Step 3: Find the \(n\)-th term of the original series
To find the \(n\)-th term of the original series \(T_n\), we can express it in terms of the first term and the sum of the differences:
\[ T_n = T_1 + \sum_{k=1}^{n-1} d_k \]
\[ T_n = 2 + \sum_{k=1}^{n-1} (2k + 3) \]

### Step 4: Calculate the sum of the differences
The sum can be split into two parts:
\[ \sum_{k=1}^{n-1} (2k + 3) = \sum_{k=1}^{n-1} 2k + \sum_{k=1}^{n-1} 3 \]
Using the formula for the sum of the first \(m\) natural numbers:
\[ \sum_{k=1}^{m} k = \frac{m(m + 1)}{2} \]
we can find:
\[ \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2} \]
Thus,
\[ \sum_{k=1}^{n-1} 2k = 2 \cdot \frac{(n-1)n}{2} = (n-1)n \]
And,
\[ \sum_{k=1}^{n-1} 3 = 3(n-1) \]

Combining these, we have:
\[ \sum_{k=1}^{n-1} (2k + 3) = (n-1)n + 3(n-1) = n^2 - n + 3n - 3 = n^2 + 2n - 3 \]

### Step 5: Substitute back to find \(T_n\)
Now substituting back into the equation for \(T_n\):
\[ T_n = 2 + (n^2 + 2n - 3) = n^2 + 2n - 1 \]

### Step 6: Find the 99th term
Now, we can find the 99th term by substituting \(n = 99\):
\[ T_{99} = 99^2 + 2 \cdot 99 - 1 \]
Calculating this:
\[ T_{99} = 9801 + 198 - 1 = 9998 \]

Thus, the 99th term of the series is \(9998\).

### Final Answer:
The 99th term of the series is \(9998\).

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