The sum of the numbers of the `n^(th)` term of the series `(1) +(1+2) +(1+2+3) +(1 +2+3+4) +… (1 +2+3+….n)`

Options

A. `""^(n+1)C_3`
B. `""^(n+1)C_2`
C. `""^(n)C_2`
D. `""^(n+2)C_3`

Answer

To find the sum of the numbers of the nth term of the series \( (1) + (1+2) + (1+2+3) + (1+2+3+4) + \ldots + (1 + 2 + 3 + \ldots + n) \), we can follow these steps:

### Step 1: Identify the nth term of the series
The nth term of the series is the sum of the first n natural numbers. The formula for the sum of the first n natural numbers is given by:
\[ S_n = \frac{n(n + 1)}{2} \]
Thus, the nth term of the series can be expressed as:
\[ T_n = S_n = \frac{n(n + 1)}{2} \]

### Step 2: Find the sum of the series up to the nth term
To find the sum of the series up to the nth term, we need to sum all the terms from \( T_1 \) to \( T_n \):
\[ \text{Sum} = T_1 + T_2 + T_3 + \ldots + T_n \]

Using the formula for \( T_k \):
\[ \text{Sum} = \sum_{k=1}^{n} \frac{k(k + 1)}{2} \]
This can be simplified as:
\[ \text{Sum} = \frac{1}{2} \sum_{k=1}^{n} k(k + 1) \]

### Step 3: Expand the summation
We can expand \( k(k + 1) \) as:
\[ k(k + 1) = k^2 + k \]
Thus, we can rewrite the summation:
\[ \sum_{k=1}^{n} k(k + 1) = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

### Step 4: Use formulas for summation
We can use the formulas for the sum of the first n natural numbers and the sum of the squares of the first n natural numbers:
- The sum of the first n natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- The sum of the squares of the first n natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Substitute the formulas back
Substituting these formulas back into our summation gives:
\[ \sum_{k=1}^{n} k(k + 1) = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]

### Step 6: Simplify the expression
To combine these two fractions, we need a common denominator, which is 6:
\[ \sum_{k=1}^{n} k(k + 1) = \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \]
\[ = \frac{n(n + 1)(2n + 1 + 3)}{6} \]
\[ = \frac{n(n + 1)(2n + 4)}{6} \]
\[ = \frac{n(n + 1)(n + 2)}{3} \]

### Step 7: Final result
Now, we can substitute this back into our sum:
\[ \text{Sum} = \frac{1}{2} \cdot \frac{n(n + 1)(n + 2)}{3} = \frac{n(n + 1)(n + 2)}{6} \]

Thus, the sum of the numbers of the nth term of the series is:
\[ \frac{n(n + 1)(n + 2)}{6} \]