Moran Rogers
06/14/2024 · Elementary School
Determine the value of each of the constants: \( A, B \) and \( C \) \[ \left(1+x-x^{2}\right)^{6}=1+A x+B x^{2}+C x^{3}+\cdots \]
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To find the constants \( A, B, \) and \( C \) in the expansion of \( (1+x-x^2)^6 \), we can use the multinomial expansion. The expression can be rewritten as:
\[
(1 + (x - x^2))^6
\]
We will expand this using the binomial theorem, which states that:
\[
(1 + u)^n = \sum_{k=0}^{n} \binom{n}{k} u^k
\]
In our case, \( u = x - x^2 \) and \( n = 6 \). Thus, we have:
\[
(1 + (x - x^2))^6 = \sum_{k=0}^{6} \binom{6}{k} (x - x^2)^k
\]
Next, we need to expand \( (x - x^2)^k \). Using the binomial theorem again, we have:
\[
(x - x^2)^k = \sum_{j=0}^{k} \binom{k}{j} x^j (-x^2)^{k-j} = \sum_{j=0}^{k} \binom{k}{j} (-1)^{k-j} x^{j + 2(k-j)} = \sum_{j=0}^{k} \binom{k}{j} (-1)^{k-j} x^{2k - j}
\]
Now, we need to find the coefficients of \( x^1, x^2, \) and \( x^3 \) in the overall expansion.
### Coefficient of \( x^1 \) (i.e., \( A \))
To get \( x^1 \), we need \( 2k - j = 1 \), which gives \( j = 2k - 1 \). This means \( j \) must be odd, and \( k \) must be at least 1 (since \( j \) cannot be negative). The possible values of \( k \) are \( 1, 2, 3, \ldots, 6 \).
- For \( k = 1 \): \( j = 1 \) gives \( \binom{1}{1} (-1)^{0} = 1 \)
- For \( k = 2 \): \( j = 3 \) gives \( \binom{2}{3} = 0 \) (not possible)
- For \( k = 3 \): \( j = 5 \) gives \( \binom{3}{5} = 0 \) (not possible)
- For \( k = 4 \): \( j = 7 \) gives \( \binom{4}{7} = 0 \) (not possible)
- For \( k = 5 \): \( j = 9 \) gives \( \binom{5}{9} = 0 \) (not possible)
- For \( k = 6 \): \( j = 11 \) gives \( \binom{6}{11} = 0 \) (not possible)
Thus, the only contribution to \( x^1 \) comes from \( k = 1 \):
\[
A = 1
\]
### Coefficient of \( x^2 \) (i.e., \( B \))
To get \( x^2 \), we need \( 2k - j = 2 \), which gives \( j = 2k - 2 \). The possible values of \( k \) are \( 1, 2, 3, \ldots, 6 \).
- For \( k = 1 \): \( j = 0 \) gives \( \binom{1}{0} (-1)^{1} = -1 \)
- For \( k = 2 \): \( j = 2 \) gives \( \binom{2}{2} (-1)^{0} = 1 \)
- For \( k = 3 \): \( j = 4 \) gives \( \binom{3}{4} = 0 \) (not possible)
- For \( k = 4 \): \( j = 6 \) gives \( \binom{4}{6} = 0 \) (not possible)
- For \( k = 5 \): \( j = 8 \) gives \( \binom{5}{8} = 0 \) (not possible)
- For \( k = 6 \): \( j = 10 \) gives \( \binom{6}{10} = 0 \) (not possible)
Thus, the contributions to \( x^2 \) are
Quick Answer
\( A = 1 \)
\( B = -1 + 1 = 0 \)
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