Show that the permutations of the letters of the word AMERICA is the same as the permutations of the letters of the word CALCUTTA. Expand (x - 2)^9 and find the sixth term using th... Show that the permutations of the letters of the word AMERICA is the same as the permutations of the letters of the word CALCUTTA. Expand (x - 2)^9 and find the sixth term using the Binomial theorem. Read more

Steps to Solve

1. Understanding the Problem
   We need to expand the expression $(x - 2)^{99}$ using the Binomial Theorem and find the sixth term in the expansion.

2. Using the Binomial Theorem
   According to the Binomial Theorem, the expansion of $(a + b)^n$ is given by:

$$ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$

For our case, we have (a = x), (b = -2), and (n = 99).

3. Identifying the Sixth Term
   The $k^{th}$ term in the expansion can be written as:

$$ T_{k+1} = \binom{n}{k} a^{n-k} b^k $$

To find the sixth term, we set (k = 5):

$$ T_6 = \binom{99}{5} x^{99-5} (-2)^5 $$

4. Calculating the Binomial Coefficient
   We calculate the binomial coefficient:

$$ \binom{99}{5} = \frac{99!}{5! \cdot (99-5)!} $$

5. Calculating the Powers
   Next, calculate the powers:

$$ x^{99-5} = x^{94} $$

And for (b^5):

$$ (-2)^5 = -32 $$

6. Putting It All Together
   Now, combine all parts together to express (T_6):

$$ T_6 = \binom{99}{5} x^{94} (-32) $$

7. Final Calculation
   Now we compute:

$$ T_6 = -32 \cdot \binom{99}{5} x^{94} $$

We can leave the answer in terms of the binomial coefficient.