Sum of the series `sum_0^10 (-1)^r 10C_r (1/3^r+8^r/3^(2r)) is `

Options

A. `(2^(10)+1)/(3^(20))`
B. `((2)/(3))^(10) - ((1)/(9))^(10)`
C. `(3^(10)-1)/(3^(20))`
D. `(6^(10)+1)/(3^(20))`

Explanation

To find the sum of the series

\[ \sum_{r=0}^{10} (-1)^r \binom{10}{r} \left( \frac{1}{3^r} + \frac{8^r}{3^{2r}} \right), \]

we can break this down into two separate sums. Let's denote the sum as \( S \):

\[ S = \sum_{r=0}^{10} (-1)^r \binom{10}{r} \frac{1}{3^r} + \sum_{r=0}^{10} (-1)^r \binom{10}{r} \frac{8^r}{3^{2r}}. \]

### Step 1: Simplifying the first sum

The first sum can be expressed as:

\[ S_1 = \sum_{r=0}^{10} (-1)^r \binom{10}{r} \frac{1}{3^r}. \]

Using the binomial theorem, we know that:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} x^r = (1-x)^n. \]

In our case, \( n = 10 \) and \( x = \frac{1}{3} \):

\[ S_1 = \left(1 - \frac{1}{3}\right)^{10} = \left(\frac{2}{3}\right)^{10}. \]

### Step 2: Simplifying the second sum

The second sum can be expressed as:

\[ S_2 = \sum_{r=0}^{10} (-1)^r \binom{10}{r} \frac{8^r}{3^{2r}} = \sum_{r=0}^{10} (-1)^r \binom{10}{r} \left(\frac{8}{9}\right)^r. \]

Again, applying the binomial theorem:

\[ S_2 = \left(1 - \frac{8}{9}\right)^{10} = \left(\frac{1}{9}\right)^{10}. \]

### Step 3: Combining the results

Now we can combine \( S_1 \) and \( S_2 \):

\[ S = S_1 + S_2 = \left(\frac{2}{3}\right)^{10} + \left(\frac{1}{9}\right)^{10}. \]

### Step 4: Finding a common denominator

To combine these fractions, we can express \( \left(\frac{1}{9}\right)^{10} \) in terms of base 3:

\[ \left(\frac{1}{9}\right)^{10} = \left(\frac{1}{3^2}\right)^{10} = \frac{1}{3^{20}}. \]

Now we have:

\[ S = \frac{2^{10}}{3^{10}} + \frac{1}{3^{20}}. \]

Finding a common denominator, we get:

\[ S = \frac{2^{10} \cdot 3^{10}}{3^{20}} + \frac{1}{3^{20}} = \frac{2^{10} \cdot 3^{10} + 1}{3^{20}}. \]

### Final Result

Thus, the sum of the series is:

\[ S = \frac{2^{10} \cdot 3^{10} + 1}{3^{20}}. \]