Find the sum of n terms of series : 3 + 5 + 9 + 15 + 23 + ………… n terms

Answer

To find the sum of the first n terms of the series: 3 + 5 + 9 + 15 + 23 + …, we can follow these steps:

### Step 1: Identify the pattern in the series
The given series is: 3, 5, 9, 15, 23, ...
To find a pattern, we can look at the differences between consecutive terms:
- 5 - 3 = 2
- 9 - 5 = 4
- 15 - 9 = 6
- 23 - 15 = 8

The differences are: 2, 4, 6, 8, which are increasing by 2 each time. This suggests that the series is generated by a quadratic function.

### Step 2: Find the nth term
Let’s denote the nth term of the series as \( a_n \). Since the differences are linear, we can assume that \( a_n \) is a quadratic function of n:
\[ a_n = An^2 + Bn + C \]

We can find A, B, and C using the first few terms:
1. For \( n = 1 \): \( a_1 = 3 \)
\[ A(1^2) + B(1) + C = 3 \]
\[ A + B + C = 3 \] (Equation 1)

2. For \( n = 2 \): \( a_2 = 5 \)
\[ A(2^2) + B(2) + C = 5 \]
\[ 4A + 2B + C = 5 \] (Equation 2)

3. For \( n = 3 \): \( a_3 = 9 \)
\[ A(3^2) + B(3) + C = 9 \]
\[ 9A + 3B + C = 9 \] (Equation 3)

### Step 3: Solve the equations
Now we have a system of three equations:
1. \( A + B + C = 3 \)
2. \( 4A + 2B + C = 5 \)
3. \( 9A + 3B + C = 9 \)

We can subtract Equation 1 from Equation 2:
\[ (4A + 2B + C) - (A + B + C) = 5 - 3 \]
\[ 3A + B = 2 \] (Equation 4)

Now subtract Equation 2 from Equation 3:
\[ (9A + 3B + C) - (4A + 2B + C) = 9 - 5 \]
\[ 5A + B = 4 \] (Equation 5)

Now we can solve Equations 4 and 5:
Subtract Equation 4 from Equation 5:
\[ (5A + B) - (3A + B) = 4 - 2 \]
\[ 2A = 2 \]
\[ A = 1 \]

Now substitute \( A = 1 \) back into Equation 4:
\[ 3(1) + B = 2 \]
\[ 3 + B = 2 \]
\[ B = -1 \]

Now substitute \( A \) and \( B \) back into Equation 1:
\[ 1 - 1 + C = 3 \]
\[ C = 3 \]

So, the nth term is:
\[ a_n = n^2 - n + 3 \]

### Step 4: Find the sum of the first n terms
The sum of the first n terms \( S_n \) is given by:
\[ S_n = a_1 + a_2 + a_3 + ... + a_n \]
\[ S_n = \sum_{k=1}^{n} (k^2 - k + 3) \]
\[ S_n = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 3 \]

Using the formulas for the sums:
- \( \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \)
- \( \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \)
- \( \sum_{k=1}^{n} 3 = 3n \)

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

### Step 5: Simplify the expression
Combine the terms:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} - \frac{3n(n + 1)}{6} + 3n \]
\[ S_n = \frac{n(n + 1)(2n + 1 - 3)}{6} + 3n \]
\[ S_n = \frac{n(n + 1)(2n - 2)}{6} + 3n \]
\[ S_n = \frac{n(n + 1)(2(n - 1))}{6} + 3n \]
\[ S_n = \frac{n(n + 1)(n - 1)}{3} + 3n \]

Thus, the final expression for the sum of the first n terms of the series is:
\[ S_n = \frac{n(n^2 - 1)}{3} + 3n \]