Find the sum of n terms of the series 1+4+10+20+35+…

Explanation

To find the sum of the first n terms of the series \(1 + 4 + 10 + 20 + 35 + \ldots\), we will first identify the pattern in the series and then derive a formula for the sum.

### Step 1: Identify the nth term of the series
The given series is:
- \(T_1 = 1\)
- \(T_2 = 4\)
- \(T_3 = 10\)
- \(T_4 = 20\)
- \(T_5 = 35\)

Let's observe the differences between consecutive terms:
- \(T_2 - T_1 = 4 - 1 = 3\)
- \(T_3 - T_2 = 10 - 4 = 6\)
- \(T_4 - T_3 = 20 - 10 = 10\)
- \(T_5 - T_4 = 35 - 20 = 15\)

Now, let's observe the second differences:
- \(6 - 3 = 3\)
- \(10 - 6 = 4\)
- \(15 - 10 = 5\)

The second differences are increasing by 1, which suggests that the nth term can be expressed as a quadratic function.

### Step 2: Formulate the nth term
Assuming \(T_n = an^2 + bn + c\), we can use the first few terms to set up equations:
1. For \(n = 1\): \(a(1)^2 + b(1) + c = 1\) → \(a + b + c = 1\) (1)
2. For \(n = 2\): \(a(2)^2 + b(2) + c = 4\) → \(4a + 2b + c = 4\) (2)
3. For \(n = 3\): \(a(3)^2 + b(3) + c = 10\) → \(9a + 3b + c = 10\) (3)

Now we have a system of equations:
- From (1): \(a + b + c = 1\)
- From (2): \(4a + 2b + c = 4\)
- From (3): \(9a + 3b + c = 10\)

### Step 3: Solve the system of equations
Subtract (1) from (2):
\[ (4a + 2b + c) - (a + b + c) = 4 - 1 \\ 3a + b = 3 \quad \text{(4)} \]

Subtract (2) from (3):
\[ (9a + 3b + c) - (4a + 2b + c) = 10 - 4 \\ 5a + b = 6 \quad \text{(5)} \]

Now, subtract (4) from (5):
\[ (5a + b) - (3a + b) = 6 - 3 \\ 2a = 3 \\ a = \frac{3}{2} \]

Substituting \(a\) back into (4):
\[ 3\left(\frac{3}{2}\right) + b = 3 \\ \frac{9}{2} + b = 3 \\ b = 3 - \frac{9}{2} = -\frac{3}{2} \]

Now substitute \(a\) and \(b\) into (1):
\[ \frac{3}{2} - \frac{3}{2} + c = 1 \\ c = 1 \]

Thus, the nth term is:
\[ T_n = \frac{3}{2}n^2 - \frac{3}{2}n + 1 \]

### Step 4: Find the sum of the first n terms
The sum of the first n terms \(S_n\) can be calculated using:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n \]

Using the formula for the sum of the first n terms of a quadratic sequence:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(\frac{3}{2}k^2 - \frac{3}{2}k + 1\right) \]

This can be separated into three sums:
\[ S_n = \frac{3}{2} \sum_{k=1}^{n} k^2 - \frac{3}{2} \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

Using the formulas:
- \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
- \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
- \(\sum_{k=1}^{n} 1 = n\)

Substituting these into the equation:
\[ S_n = \frac{3}{2} \cdot \frac{n(n+1)(2n+1)}{6} - \frac{3}{2} \cdot \frac{n(n+1)}{2} + n \]

### Step 5: Simplify the expression
\[ S_n = \frac{3n(n+1)(2n+1)}{12} - \frac{3n(n+1)}{4} + n \]

Finding a common denominator (12):
\[ S_n = \frac{3n(n+1)(2n+1) - 9n(n+1) + 12n}{12} \]

Now, simplify the numerator:
\[ 3n(n+1)(2n+1) - 9n(n+1) + 12n = n(n+1)(3(2n+1) - 9) + 12n \]
\[ = n(n+1)(6n - 6) + 12n = 6n(n+1)(n-1) + 12n \]

Thus, the final expression for \(S_n\) is:
\[ S_n = \frac{n(n+1)(n+2)}{6} \]

### Final Answer
The sum of the first n terms of the series \(1 + 4 + 10 + 20 + 35 + \ldots\) is:
\[ S_n = \frac{n(n+1)(n+2)}{6} \]