Find the sum to n terms of the series `3+6 + 10 + 16` + …

Options

A. `(n(n-1))/(2) -1`
B. `n(n+1) + 2^n -1`
C. `n(n+1) + 1`
D. `3(2n+1)-2^n`

Answer

To find the sum to n terms of the series \(3 + 6 + 10 + 16 + \ldots\), we first need to identify the pattern in the series.

### Step 1: Identify the pattern in the series
Let's look at the series:
- The first term is \(3\).
- The second term is \(6\).
- The third term is \(10\).
- The fourth term is \(16\).

To find the differences between consecutive terms:
- \(6 - 3 = 3\)
- \(10 - 6 = 4\)
- \(16 - 10 = 6\)

Now, let's find the second differences:
- \(4 - 3 = 1\)
- \(6 - 4 = 2\)

The second differences are not constant, but the first differences are increasing. This suggests that the series might be quadratic in nature.

### Step 2: Form a quadratic equation
Assume the nth term of the series can be expressed as:
\[ a_n = An^2 + Bn + C \]
We need to find the values of \(A\), \(B\), and \(C\) using the first few terms.

Using the first three terms:
1. For \(n=1\): \(A(1^2) + B(1) + C = 3\) → \(A + B + C = 3\) (Equation 1)
2. For \(n=2\): \(A(2^2) + B(2) + C = 6\) → \(4A + 2B + C = 6\) (Equation 2)
3. For \(n=3\): \(A(3^2) + B(3) + C = 10\) → \(9A + 3B + C = 10\) (Equation 3)

### Step 3: Solve the equations
Now we solve these three equations:
1. From Equation 1: \(C = 3 - A - B\)
2. Substitute \(C\) in Equation 2:
\[ 4A + 2B + (3 - A - B) = 6 \]
Simplifying gives:
\[ 3A + B + 3 = 6 \implies 3A + B = 3 \quad (Equation 4) \]

3. Substitute \(C\) in Equation 3:
\[ 9A + 3B + (3 - A - B) = 10 \]
Simplifying gives:
\[ 8A + 2B + 3 = 10 \implies 8A + 2B = 7 \implies 4A + B = \frac{7}{2} \quad (Equation 5) \]

### Step 4: Solve Equations 4 and 5
Now we have:
1. \(3A + B = 3\) (Equation 4)
2. \(4A + B = \frac{7}{2}\) (Equation 5)

Subtract Equation 4 from Equation 5:
\[ (4A + B) - (3A + B) = \frac{7}{2} - 3 \]
This simplifies to:
\[ A = \frac{1}{2} \]

Substituting \(A = \frac{1}{2}\) back into Equation 4:
\[ 3\left(\frac{1}{2}\right) + B = 3 \implies \frac{3}{2} + B = 3 \implies B = \frac{3}{2} \]

Now substituting \(A\) and \(B\) back into Equation 1 to find \(C\):
\[ \frac{1}{2} + \frac{3}{2} + C = 3 \implies 2 + C = 3 \implies C = 1 \]

### Step 5: Write the nth term
Thus, the nth term of the series is:
\[ a_n = \frac{1}{2}n^2 + \frac{3}{2}n + 1 \]

### Step 6: Find the sum of n terms
The sum \(S_n\) of the first \(n\) terms is given by:
\[ S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} \left(\frac{1}{2}k^2 + \frac{3}{2}k + 1\right) \]
Using the formulas for the sums:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} 1 = n \]

We can calculate:
\[ S_n = \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{3}{2} \cdot \frac{n(n+1)}{2} + n \]

### Step 7: Simplify the sum
Combining these gives:
\[ S_n = \frac{n(n+1)(2n+1)}{12} + \frac{3n(n+1)}{4} + n \]
To combine these fractions, find a common denominator (12):
\[ S_n = \frac{n(n+1)(2n+1)}{12} + \frac{9n(n+1)}{12} + \frac{12n}{12} \]
\[ S_n = \frac{n(n+1)(2n+1) + 9n(n+1) + 12n}{12} \]
\[ S_n = \frac{n(n+1)(2n + 10) + 12n}{12} \]

This gives the final expression for the sum of the first \(n\) terms of the series.