Key Concept: Sum of Coefficients of Even Powers

Let a polynomial be expressed in its general form as $P(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_k x^k$. The sum of the coefficients of all even powers of $x$ (i.e., $a_0 + a_2 + a_4 + \dots$) in $P(x)$ is given by the formula: $\text{Sum of coefficients of even powers} = \frac{P(1) + P(-1)}{2}$

Why this formula works:

1. Evaluating at $x=1$: When we substitute $x=1$ into the polynomial, all powers of $1$ are $1$. This gives us the sum of all coefficients: $P(1) = a_0 + a_1(1) + a_2(1)^2 + a_3(1)^3 + \dots = a_0 + a_1 + a_2 + a_3 + \dots \quad \text{(Equation 1)}$
2. Evaluating at $x=-1$: When we substitute $x=-1$ into the polynomial, the odd powers of $x$ become $-1$, and the even powers become $1$. This results in an alternating sum of coefficients: $P(-1) = a_0 + a_1(-1) + a_2(-1)^2 + a_3(-1)^3 + \dots = a_0 - a_1 + a_2 - a_3 + \dots \quad \text{(Equation 2)}$
3. Adding the Equations: Adding Equation 1 and Equation 2, the terms with odd-indexed coefficients ($a_1, a_3, \dots$) cancel each other out, while the terms with even-indexed coefficients ($a_0, a_2, \dots$) are doubled: $P(1) + P(-1) = (a_0 + a_1 + a_2 + a_3 + \dots) + (a_0 - a_1 + a_2 - a_3 + \dots)$ $P(1) + P(-1) = 2a_0 + 2a_2 + 2a_4 + \dots$ $P(1) + P(-1) = 2(a_0 + a_2 + a_4 + \dots)$ Dividing by $2$ isolates the sum of coefficients of even powers, thus proving the formula.

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Step 1: Define the Polynomials

We are given the product of two polynomials: $P(x) = (1 + x + x^2 + \dots + x^{2n})(1 - x + x^2 - x^3 + \dots + x^{2n})$ To simplify our calculations, let's denote the first polynomial as $A(x)$ and the second as $B(x)$: $A(x) = 1 + x + x^2 + \dots + x^{2n}$ $B(x) = 1 - x + x^2 - x^3 + \dots + x^{2n}$ So, $P(x) = A(x) \cdot B(x)$. Our goal is to find $n$ given that the sum of coefficients of even powers of $x$ in $P(x)$ is $61$.

Why this step is important: Breaking down the complex product into two simpler polynomials, $A(x)$ and $B(x)$, makes the evaluation process much clearer and reduces the chance of errors.

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Step 2: Evaluate $P(1)$

To find $P(1)$, we first evaluate $A(1)$ and $B(1)$ and then multiply them.

1. Evaluate $A(1)$: Substitute $x=1$ into $A(x)$: $A(1) = 1 + 1 + 1^2 + \dots + 1^{2n}$ $A(1) = 1 + 1 + 1 + \dots + 1 \quad (\text{There are } 2n+1 \text{ terms from } x^0 \text{ to } x^{2n})$ $A(1) = 2n + 1$
2. Evaluate $B(1)$: Substitute $x=1$ into $B(x)$: $B(1) = 1 - 1 + 1^2 - 1^3 + \dots + 1^{2n}$ $B(1) = 1 - 1 + 1 - 1 + \dots + 1$ Since there are $2n+1$ terms (an odd number of terms) and they alternate between $+1$ and $-1$, the sum will simplify to $1$. For example, $(1-1) + (1-1) + \dots + (1-1) + 1 = 1$. $B(1) = 1$
3. Calculate $P(1)$: $P(1) = A(1) \cdot B(1) = (2n + 1) \cdot 1$ $P(1) = 2n + 1$

Why this step is taken: $P(1)$ represents the sum of all coefficients of $P(x)$. This is the first crucial component required by our main formula. Carefully evaluating the sum of the terms for $A(1)$ and $B(1)$ is essential.

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Step 3: Evaluate $P(-1)$

Next, we evaluate $A(-1)$ and $B(-1)$ and then multiply them.

1. Evaluate $A(-1)$: Substitute $x=-1$ into $A(x)$: $A(-1) = 1 + (-1) + (-1)^2 + \dots + (-1)^{2n}$ $A(-1) = 1 - 1 + 1 - 1 + \dots + (-1)^{2n}$ Since $2n$ is an even number, $(-1)^{2n} = 1$. The series has $2n+1$ terms (an odd number). Similar to $B(1)$, the sum will simplify to $1$. $A(-1) = 1$
2. Evaluate $B(-1)$: Substitute $x=-1$ into $B(x)$: $B(-1) = 1 - (-1) + (-1)^2 - (-1)^3 + \dots + (-1)^{2n}$ $B(-1) = 1 + 1 + 1 + 1 + \dots + 1$ Here, each term becomes positive. For example, $-(-1)^3 = -(-1) = 1$. All $2n+1$ terms are $1$. $B(-1) = 2n + 1$
3. Calculate $P(-1)$: $P(-1) = A(-1) \cdot B(-1) = 1 \cdot (2n + 1)$ $P(-1) = 2n + 1$

Why this step is taken: $P(-1)$ gives us the alternating sum of coefficients of $P(x)$. This is the second crucial component for our main formula, allowing us to isolate the even-powered coefficients. Paying careful attention to the signs when $x=-1$ is critical.

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Step 4: Calculate the Sum of Coefficients of Even Powers

Now, we apply the formula from the Key Concept section using the values we found for $P(1)$ and $P(-1)$: $\text{Sum of coefficients of even powers} = \frac{P(1) + P(-1)}{2}$ Substitute $P(1) = 2n+1$ and $P(-1) = 2n+1$: $\text{Sum} = \frac{(2n + 1) + (2n + 1)}{2}$ $\text{Sum} = \frac{2(2n + 1)}{2}$ $\text{Sum} = 2n + 1$

Why this step is taken: This step directly uses the proven algebraic property to combine the results from $P(1)$ and $P(-1)$ to arrive at an expression for the desired sum in terms of $n$.

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Step 5: Solve for n

The problem states that the sum of the coefficients of all even powers of $x$ in the product is $61$. From Step 4, we derived this sum to be $2n + 1$. Therefore, we can set up the following equation: $2n + 1 = 61$ Now, we solve this linear equation for $n$: Subtract $1$ from both sides: $2n = 61 - 1$ $2n = 60$ Divide both sides by $2$: $n = \frac{60}{2}$ $n = 30$

Why this step is taken: This is the final algebraic step to determine the specific value of $n$ by using the information provided in the question.

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Tips for Success & Common Mistakes

• Geometric Series Summation: Both $A(x)$ and $B(x)$ are finite geometric series. A series $1+r+r^2+\dots+r^k$ has $k+1$ terms. For $A(x)$, $k=2n$, so there are $2n+1$ terms. This count is vital when evaluating sums like $1-1+1-\dots$.
• Sign Management for $x=-1$: When substituting $x=-1$, be extremely careful with signs. Remember that $(-1)^{\text{even power}} = 1$ and $(-1)^{\text{odd power}} = -1$. In $B(x) = 1 - x + x^2 - x^3 + \dots$, the terms evaluated at $x=-1$ become $1 - (-1) + (-1)^2 - (-1)^3 + \dots = 1 + 1 + 1 + 1 + \dots$.
• General Formula for Odd Powers: Similarly, the sum of coefficients of odd powers ($a_1 + a_3 + \dots$) is given by $\frac{P(1) - P(-1)}{2}$. It's useful to remember both formulas.
• Not Expanding Explicitly: The beauty of this method is that it avoids the tedious and error-prone process of fully expanding the product of two polynomials, which would be very complex for $x^{2n}$.

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Summary

We leveraged the fundamental property of polynomials that allows us to find the sum of coefficients of even powers by calculating $P(1)$ and $P(-1)$. By evaluating the individual polynomial factors $A(x)$ and $B(x)$ at $x=1$ and $x=-1$, we found $P(1) = 2n+1$ and $P(-1) = 2n+1$. Applying the formula $\frac{P(1) + P(-1)}{2}$, we determined the sum of coefficients of even powers to be $2n+1$. Finally, by equating this to the given value of $61$, we solved the simple linear equation to find $n=30$. This technique provides an elegant and efficient solution to a problem that would otherwise involve extensive polynomial multiplication.