`sum_(r=1)^n (a^r +br) , a , b in R^+` is equal to :

Options

A. `(1-a^n)/(1-a) + (bn(n+1))/2`
B. `(b(1-a^n))/(1-a) + (an(n+1))/2`
C. `(a(1-a^n))/(1-a) + (bn(n+1))/2`
D. `(a(1-a^n))/(1-a) + (an(n+1))/2`

Answer

To solve the summation \( \sum_{r=1}^{n} (a^r + br) \), where \( a, b \in \mathbb{R}^+ \), we can break it down into two separate summations:

1. **Summation of \( a^r \)** from \( r=1 \) to \( n \)
2. **Summation of \( br \)** from \( r=1 \) to \( n \)

### Step 1: Calculate the summation of \( a^r \)

The first part of the summation is:

\[ \sum_{r=1}^{n} a^r \]

This is a geometric series where the first term \( a^1 = a \) and the common ratio \( r = a \). The formula for the sum of the first \( n \) terms of a geometric series is given by:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

In our case, substituting \( a \) for the first term and \( a \) for the common ratio, we have:

\[ \sum_{r=1}^{n} a^r = a \frac{1 - a^n}{1 - a} \]

### Step 2: Calculate the summation of \( br \)

The second part of the summation is:

\[ \sum_{r=1}^{n} br \]

We can factor out \( b \):

\[ \sum_{r=1}^{n} br = b \sum_{r=1}^{n} r \]

The sum of the first \( n \) natural numbers is given by the formula:

\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]

Thus, substituting this into our equation gives:

\[ \sum_{r=1}^{n} br = b \cdot \frac{n(n + 1)}{2} \]

### Step 3: Combine the results

Now, we can combine both parts of the summation:

\[ \sum_{r=1}^{n} (a^r + br) = \sum_{r=1}^{n} a^r + \sum_{r=1}^{n} br \]

Substituting the results we calculated:

\[ = a \frac{1 - a^n}{1 - a} + b \cdot \frac{n(n + 1)}{2} \]

### Final Answer

Thus, the final result for the summation \( \sum_{r=1}^{n} (a^r + br) \) is:

\[ \sum_{r=1}^{n} (a^r + br) = a \frac{1 - a^n}{1 - a} + \frac{b n(n + 1)}{2} \]

---