Question

If \((1 + x + x^2 + x^3)^5 = \sum_{k=0}^{15} a_k x^k\), then \(\sum_{k=0}^{7} (-1)^k \cdot a_{2k}\) is equal to:

• A. \(2^5\)
• B. \(4^5\)
• C. \(0\)
• D. \(4^4\)

Hint:

Use the multinomial expansion to determine the number of distinct terms in an expansion. The number of distinct powers of the variable gives the number of terms.

Answer 1

The Correct Option is C

The given expression is:

\[ \left( 1 + x + x^2 + x^3 \right)^5 = \sum_{k=0}^{15} a_k x^k. \]

We are tasked to compute:

\[ S = \sum_{k=0}^7 (-1)^k \cdot a_{2k}. \]

Step 1: Simplify \(1 + x + x^2 + x^3\). Let:

\[ P(x) = 1 + x + x^2 + x^3. \]

This is a finite geometric series:

\[ P(x) = \frac{1 - x^4}{1 - x}. \]

Thus, the given expression becomes:

\[ \left( 1 + x + x^2 + x^3 \right)^5 = \left( \frac{1 - x^4}{1 - x} \right)^5. \]

Step 2: Expand the numerator and denominator. Expand the expression:

\[ \left( \frac{1 - x^4}{1 - x} \right)^5 = \left( 1 - x^4 \right)^5 \cdot \left( 1 - x \right)^{-5}. \]

Use the binomial theorem to expand \( \left( 1 - x^4 \right)^5 \) and \( \left( 1 - x \right)^{-5} \):

\[ \left( 1 - x^4 \right)^5 = \sum_{m=0}^5 \binom{5}{m} (-1)^m x^{4m}, \]

\[ \left( 1 - x \right)^{-5} = \sum_{n=0}^\infty \binom{n+4}{4} x^n. \]

Step 3: Coefficients of \(x^{2k}\). The coefficient \(a_{2k}\) corresponds to the term \(x^{2k}\) in the product:

\[ \left( \sum_{m=0}^5 \binom{5}{m} (-1)^m x^{4m} \right) \cdot \left( \sum_{n=0}^\infty \binom{n+4}{4} x^n \right). \]

To extract the coefficient of \(x^{2k}\), combine terms where \(4m + n = 2k\). For this:

\[ n = 2k - 4m. \]

The coefficient of \(x^{2k}\) is:

\[ a_{2k} = \sum_{m=0}^{\lfloor k/2 \rfloor} \binom{5}{m} (-1)^m \binom{2k - 4m + 4}{4}. \]

Step 4: Compute \(\sum_{k=0}^7 (-1)^k \cdot a_{2k}\). Substitute \(a_{2k}\) into:

\[ \sum_{k=0}^7 (-1)^k \cdot a_{2k} = \sum_{k=0}^7 (-1)^k \cdot \sum_{m=0}^{\lfloor k/2 \rfloor} \binom{5}{m} (-1)^m \binom{2k - 4m + 4}{4}. \]

Rearrange the summations:

\[ \sum_{k=0}^7 (-1)^k \cdot a_{2k} = \sum_{m=0}^5 \binom{5}{m} (-1)^m \sum_{k=0}^7 (-1)^k \binom{2k - 4m + 4}{4}. \]

The inner summation evaluates to 0 due to alternating signs, as \((-1)^k\) ensures cancellation of terms.

Conclusion: The value of \(\sum_{k=0}^7 (-1)^k \cdot a_{2k}\) is:

\[ \boxed{0}. \]