Find the sum of series `1+5+14+30+55+`…………..upto n terms.

Answer

To find the sum of the series \( S_n = 1 + 5 + 14 + 30 + 55 + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Identify the pattern in the series
The given series is \( 1, 5, 14, 30, 55, \ldots \). We need to find a formula for the \( n \)-th term of the series.

### Step 2: Find the \( n \)-th term
Let's denote the \( n \)-th term as \( T_n \). We can observe the differences between consecutive terms:
- \( 5 - 1 = 4 \)
- \( 14 - 5 = 9 \)
- \( 30 - 14 = 16 \)
- \( 55 - 30 = 25 \)

The differences are \( 4, 9, 16, 25 \), which are \( 2^2, 3^2, 4^2, 5^2 \). This suggests that the \( n \)-th term can be expressed in terms of squares.

### Step 3: Formulate the \( n \)-th term
We can express the \( n \)-th term as:
\[ T_n = T_{n-1} + n^2 \]
Starting from \( T_1 = 1 \), we can derive:
- \( T_2 = T_1 + 2^2 = 1 + 4 = 5 \)
- \( T_3 = T_2 + 3^2 = 5 + 9 = 14 \)
- \( T_4 = T_3 + 4^2 = 14 + 16 = 30 \)
- \( T_5 = T_4 + 5^2 = 30 + 25 = 55 \)

### Step 4: Sum of the series
The sum of the first \( n \) terms can be expressed as:
\[ S_n = \sum_{k=1}^{n} T_k \]
Using the formula for the sum of squares, we can write:
\[ S_n = \sum_{k=1}^{n} (1 + 2^2 + 3^2 + \ldots + n^2) \]
The formula for the sum of the first \( n \) squares is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Thus, we can express \( S_n \) as:
\[ S_n = n + \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Final expression for \( S_n \)
Combining the terms, we get:
\[ S_n = \frac{6n + n(n + 1)(2n + 1)}{6} \]
This can be simplified further, but we can also express it in a more compact form:
\[ S_n = \frac{n(n + 1)(n + 2)}{6} \]

### Final Answer
The sum of the series \( 1 + 5 + 14 + 30 + 55 + \ldots \) up to \( n \) terms is:
\[ S_n = \frac{n(n + 1)(n + 2)}{6} \]