Find the sum to n terms of the series whose nth term is
n (n+2)

Answer

To find the sum to n terms of the series whose nth term is given by \( T_n = n(n + 2) \), we can follow these steps:

### Step 1: Expand the nth term
The nth term is given as:
\[ T_n = n(n + 2) = n^2 + 2n \]

### Step 2: Write the sum of the first n terms
The sum of the first n terms \( S_n \) can be expressed as:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + 2k) \]

### Step 3: Split the sum
We can split the sum into two separate sums:
\[ S_n = \sum_{k=1}^{n} k^2 + 2\sum_{k=1}^{n} k \]

### Step 4: Use the formulas for the sums
We use the formulas for the sums of squares and the sums of the first n natural numbers:
- The sum of the first n squares is given by:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
- The sum of the first n natural numbers is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]

### Step 5: Substitute the formulas into the sum
Substituting these formulas into our expression for \( S_n \):
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + 2 \cdot \frac{n(n + 1)}{2} \]

### Step 6: Simplify the expression
Now, simplify the second term:
\[ 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]
Thus, we have:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + n(n + 1) \]

### Step 7: Combine the terms
To combine the terms, we need a common denominator:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{6n(n + 1)}{6} \]
This gives:
\[ S_n = \frac{n(n + 1)(2n + 1 + 6)}{6} = \frac{n(n + 1)(2n + 7)}{6} \]

### Final Result
Thus, the sum to n terms of the series is:
\[ S_n = \frac{n(n + 1)(2n + 7)}{6} \]

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