Find the sum to n terms of the series
`1+ (1+2) + (1+2+3) + (1+2+ 3+ 4) + ….:`

Options

A. `(n(n+1)(2n+1))/(6)`
B. `(n(n+1)(n+2))/(6)`
C. `(n(n+1)(n+2))/(12)`
D. `(n(n+1))/(2)`

Answer

To find the sum to n terms of the series \( S_n = 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + \ldots \), we can break down the problem step by step.

### Step 1: Identify the pattern in the series
The series can be rewritten as:
- The first term is \( 1 \)
- The second term is \( 1 + 2 \)
- The third term is \( 1 + 2 + 3 \)
- The fourth term is \( 1 + 2 + 3 + 4 \)
- And so on...

The k-th term of the series can be expressed as:
\[ T_k = 1 + 2 + 3 + \ldots + k = \frac{k(k + 1)}{2} \]
This formula gives us the sum of the first k natural numbers.

### Step 2: Write the sum of the first n terms
Now, we need to find the sum of the first n terms:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n \]
Substituting the formula for \( T_k \):
\[ S_n = \frac{1(1 + 1)}{2} + \frac{2(2 + 1)}{2} + \frac{3(3 + 1)}{2} + \ldots + \frac{n(n + 1)}{2} \]
This simplifies to:
\[ S_n = \frac{1}{2} \left( 1(1 + 1) + 2(2 + 1) + 3(3 + 1) + \ldots + n(n + 1) \right) \]

### Step 3: Simplify the expression
Now we need to simplify the sum:
\[ S_n = \frac{1}{2} \left( \sum_{k=1}^{n} k(k + 1) \right) \]
We can separate the terms:
\[ \sum_{k=1}^{n} k(k + 1) = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]
Using the formulas for the sums:
- \( \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \)
- \( \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \)

### Step 4: Substitute the formulas into the sum
Substituting these into our expression gives:
\[ \sum_{k=1}^{n} k(k + 1) = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]
To combine these, we need a common denominator of 6:
\[ = \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \]
This simplifies to:
\[ = \frac{n(n + 1)(2n + 1 + 3)}{6} = \frac{n(n + 1)(2n + 4)}{6} = \frac{n(n + 1)(2(n + 2))}{6} \]

### Step 5: Final expression for \( S_n \)
Now substituting back into \( S_n \):
\[ S_n = \frac{1}{2} \cdot \frac{n(n + 1)(2(n + 2))}{6} = \frac{n(n + 1)(n + 2)}{6} \]

### Final Result
Thus, the sum to n terms of the series is:
\[ S_n = \frac{n(n + 1)(n + 2)}{6} \]