The nth term of a series of which all the terms are positive is defined as `T_n = n^2 + n` then the sum of n terms of the series is :

Options

A. a. `(n (n + 1) (n + 2))/6`
B. b. `(n^3 + 4n)/n`
C. c. `(5n)/9 [n^2 + n]`
D. d. `(n(n + 1)(n +2))/3`

Answer

To find the sum of the first n terms of the series defined by the nth term \( T_n = n^2 + n \), we can follow these steps:

### Step 1: Write the sum of the first n terms
The sum of the first n terms \( S_n \) can be expressed as:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n \]
This can be rewritten using the formula for \( T_n \):
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k) \]

### Step 2: Separate the summation
We can separate the summation into two parts:
\[ S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

### Step 3: Use the formulas for the sums
We know the formulas for these summations:
1. The sum of the first n squares:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
2. The sum of the first n natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]

### Step 4: Substitute the formulas into the sum
Now we can substitute these formulas back into our expression for \( S_n \):
\[ S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]

### Step 5: Find a common denominator
To combine these two fractions, we need a common denominator. The common denominator is 6:
\[ S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} \]

### Step 6: Combine the fractions
Now we can combine the fractions:
\[ S_n = \frac{n(n+1)(2n+1 + 3)}{6} = \frac{n(n+1)(2n+4)}{6} \]

### Step 7: Simplify the expression
We can factor out a 2 from \( (2n + 4) \):
\[ S_n = \frac{n(n+1) \cdot 2(n+2)}{6} = \frac{n(n+1)(n+2)}{3} \]

### Final Result
Thus, the sum of the first n terms of the series is:
\[ S_n = \frac{n(n+1)(n+2)}{3} \]

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