If the sum of the coefficients in the expansion of `(lambda x^(2)-2x+1)^(37)` is equal to the sum of the coefficients in the expansion of `(x-lambday)^(37)`, then `lambda = `

Options

A. 0
B. 1
C. may be any real number
D. no such value exists

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the sum of the coefficients in the expansion of \( (\lambda x^2 - 2x + 1)^{37} \) is equal to the sum of the coefficients in the expansion of \( (x - \lambda y)^{37} \).

### Step 1: Find the sum of the coefficients for \( (\lambda x^2 - 2x + 1)^{37} \)

To find the sum of the coefficients in a polynomial, we substitute \( x = 1 \):

\[ \text{Sum of coefficients} = \lambda(1)^2 - 2(1) + 1 = \lambda - 2 + 1 = \lambda - 1 \]

Thus, the sum of the coefficients in the expansion of \( (\lambda x^2 - 2x + 1)^{37} \) is:

\[ (\lambda - 1)^{37} \]

### Step 2: Find the sum of the coefficients for \( (x - \lambda y)^{37} \)

Similarly, we substitute \( x = 1 \) and \( y = 1 \):

\[ \text{Sum of coefficients} = (1 - \lambda(1))^{37} = (1 - \lambda)^{37} \]

### Step 3: Set the two sums equal to each other

According to the problem, we have:

\[ (\lambda - 1)^{37} = (1 - \lambda)^{37} \]

### Step 4: Simplify the equation

Taking the 37th root of both sides (since both sides are raised to the same power, we can take the root):

\[ \lambda - 1 = 1 - \lambda \quad \text{or} \quad \lambda - 1 = -(1 - \lambda) \]

### Step 5: Solve the first case

From the first case:

\[ \lambda - 1 = 1 - \lambda \]
\[ \lambda + \lambda = 1 + 1 \]
\[ 2\lambda = 2 \]
\[ \lambda = 1 \]

### Step 6: Solve the second case

From the second case:

\[ \lambda - 1 = -1 + \lambda \]
This simplifies to \( -1 = -1 \), which is always true and does not provide any new information.

### Conclusion

Thus, the only solution we have is:

\[ \lambda = 1 \]

### Final Answer

\(\lambda = 1\)

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