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Consider the parameterized polynomial product fn(x)=(x−1)(1+x+x2+x3+⋯+xn−1)−xnf_n(x) = (x - 1)\left(1 + x + x^2 + x^3 + \dots + x^{n-1}\right) - x^n. Evaluate fn(x)f_n(x) in closed polynomial form, and determine the limit of the degree of gn(x)=(x2+2x−1)n−x2ng_n(x) = (x^2 + 2x - 1)^n - x^{2n} as a function of n≥1n \ge 1.

Answer and steps

Final answer

fn(x)=−1f_n(x) = -1 for all n≥1n \ge 1, and deg⁡(gn(x))=2n−1\deg(g_n(x)) = 2n - 1

Steps

Part 1: Evaluating fn(x)f_n(x)

To find the closed form of fn(x)=(x−1)(1+x+x2+⋯+xn−1)−xnf_n(x) = (x - 1)\left(1 + x + x^2 + \dots + x^{n-1}\right) - x^n, expand the product:

  1. Multiply the polynomial sum by xx: x⋅(1+x+x2+⋯+xn−1)=x+x2+x3+⋯+xnx \cdot \left(1 + x + x^2 + \dots + x^{n-1}\right) = x + x^2 + x^3 + \dots + x^n
  2. Multiply the polynomial sum by −1-1: −1⋅(1+x+x2+⋯+xn−1)=−1−x−x2−⋯−xn−1-1 \cdot \left(1 + x + x^2 + \dots + x^{n-1}\right) = -1 - x - x^2 - \dots - x^{n-1}
  3. Add the two distributions together to evaluate (x−1)(1+x+x2+⋯+xn−1)(x - 1)\left(1 + x + x^2 + \dots + x^{n-1}\right): (x+x2+⋯+xn)−(1+x+x2+⋯+xn−1)=xn−1(x + x^2 + \dots + x^n) - (1 + x + x^2 + \dots + x^{n-1}) = x^n - 1 Notice that all intermediate terms from xx up to xn−1x^{n-1} cancel in pairs (a telescoping sum).
  4. Substitute this result back into fn(x)f_n(x): fn(x)=(xn−1)−xn=−1f_n(x) = (x^n - 1) - x^n = -1

Thus, fn(x)=−1f_n(x) = -1 for all integers n≥1n \ge 1.

Part 2: Determining the Degree of gn(x)g_n(x)

We are given the polynomial: gn(x)=(x2+2x−1)n−x2ng_n(x) = (x^2 + 2x - 1)^n - x^{2n}

  1. Group the terms inside the base as x2+(2x−1)x^2 + (2x - 1) and expand using the Binomial Theorem: (x2+(2x−1))n=∑k=0n(nk)(x2)n−k(2x−1)k(x^2 + (2x - 1))^n = \sum_{k=0}^n \binom{n}{k} (x^2)^{n-k} (2x - 1)^k

  2. Write out the two terms of highest degree corresponding to k=0k = 0 and k=1k = 1:

    • For k=0k = 0: (n0)(x2)n(2x−1)0=x2n\binom{n}{0}(x^2)^n (2x - 1)^0 = x^{2n}
    • For k=1k = 1: (n1)(x2)n−1(2x−1)1=nx2n−2(2x−1)=2nx2n−1−nx2n−2\binom{n}{1}(x^2)^{n-1}(2x - 1)^1 = n x^{2n-2}(2x - 1) = 2n x^{2n-1} - n x^{2n-2}
    • For k≥2k \ge 2: The maximum degree of each term is 2(n−k)+k=2n−k≤2n−22(n - k) + k = 2n - k \le 2n - 2.
  3. Substitute the expansion into gn(x)g_n(x): gn(x)=[x2n+2nx2n−1+O(x2n−2)]−x2ng_n(x) = \left[x^{2n} + 2n x^{2n-1} + \mathcal{O}(x^{2n-2})\right] - x^{2n} gn(x)=2nx2n−1+O(x2n−2)g_n(x) = 2n x^{2n-1} + \mathcal{O}(x^{2n-2})
  4. The highest power of xx remaining is 2n−12n - 1, and its leading coefficient is 2n2n. Since n≥1n \ge 1, 2n≠02n \ne 0, meaning the leading term does not vanish.

Therefore, deg⁡(gn(x))=2n−1\deg(g_n(x)) = 2n - 1.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misidentifies the telescoping product (x - 1)(1 + x + ... + xⁿ⁻¹) as 1 - xⁿ instead of xⁿ - 1, concluding that f_n(x) = 1 - 2xⁿ.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    (x - 1)(1 + x + x^2 + …\dots + x^{n-1}) = 1 - x^n

    Correct

    (x - 1)(1 + x + x^2 + …\dots + x^{n-1}) = x^n - 1

    Multiplying by xx gives the positive leading term +xn+x^n, while multiplying by −1-1 gives the constant term −1-1. The identity with (x−1)(x - 1) expands to xn−1x^n - 1, whereas (1−x)(1 - x) expands to 1−xn1 - x^n.

  2. Step 2

    f_n(x) = 1 - x^n - x^n = 1 - 2x^n

    Correct

    f_n(x) = (x^n - 1) - x^n = -1

    Because of the reversed sign in the previous expansion, subtracting xnx^n led to 1−2xn1 - 2x^n instead of (xn−1)−xn=−1(x^n - 1) - x^n = -1.

  3. Step 3

    (x^2 + 2x - 1)^n = x^{2n} + n(x^2)^{n-1}(2x) + …\dots = x^{2n} + 2nx^{2n-1} + …\dots   ⟹  \implies g_n(x) = 2nx^{2n-1} + …\dots

  4. Step 4

    deg⁡(gn(x))=2n−1  ⟹  lim⁡(deg⁡(gn(x)))=∞\deg(g_n(x)) = 2n - 1 \implies \lim(\deg(g_n(x))) = \infty

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly applied the Binomial Theorem to find the first two leading terms of (x2+2x−1)n(x^2 + 2x - 1)^n.

You canceled the dominant x2nx^{2n} term to correctly determine that deg⁡(gn(x))=2n−1\deg(g_n(x)) = 2n - 1.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign error in geometric series product

You swapped the order of terms to 1−xn1 - x^n rather than xn−1x^n - 1 when expanding (x−1)(1+x+⋯+xn−1)(x - 1)(1 + x + \dots + x^{n-1}).

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the sign convention of the telescoping factorization: (1−x)(1+x+⋯+xn−1)=1−xn(1 - x)(1 + x + \dots + x^{n-1}) = 1 - x^n, while (x−1)(1+x+⋯+xn−1)=xn−1(x - 1)(1 + x + \dots + x^{n-1}) = x^n - 1.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the highest-degree product

  • The highest power comes from x⋅xn−1=+xnx \cdot x^{n-1} = +x^n, so the leading term must have a positive sign.

  • Always multiply the highest-degree terms first to verify whether the leading coefficient should be +1+1 or −1-1.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand (x−1)(1+x+x2+⋯+xn−1)=x(1+x+⋯+xn−1)−1(1+x+⋯+xn−1)=(x+x2+⋯+xn)−(1+x+⋯+xn−1)=xn−1(x - 1)(1 + x + x^2 + \dots + x^{n-1}) = x(1 + x + \dots + x^{n-1}) - 1(1 + x + \dots + x^{n-1}) = (x + x^2 + \dots + x^n) - (1 + x + \dots + x^{n-1}) = x^n - 1.

  2. Substitute this back into fn(x)f_n(x): fn(x)=(xn−1)−xn=−1f_n(x) = (x^n - 1) - x^n = -1.

Final answer

f_n(x) = -1  for all \text{ for all } n ≥\ge 1,  and \text{ and } \deg(g_n(x)) = 2n - 1

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the signs of the two classic factoring patterns: xn−1=(x−1)(xn−1+⋯+1)x^n - 1 = (x - 1)(x^{n-1} + \dots + 1) vs. 1−xn=(1−x)(1+x+⋯+xn−1)1 - x^n = (1 - x)(1 + x + \dots + x^{n-1}).

  • Test with a small value like n=2n = 2: (x−1)(1+x)=x2−1(x - 1)(1 + x) = x^2 - 1, not 1−x21 - x^2.

OverviewACEpath Teacher's Diagnosis

02 The student assumes that subtracting x²ⁿ removes all terms involving x entirely or misidentifies the leading non-canceling term as having degree 2n - 2 instead of 2n - 1 by ignoring the linear term in the binomial expansion.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    (x - 1)(1 + x + x^2 + …\dots + x^{n-1}) = x^n - 1   ⟹  \implies f_n(x) = (x^n - 1) - x^n = -1

  2. Step 2

    (x^2 + 2x - 1)^n = (x^2)^n + (2x - 1)^n = x^{2n} + (2x - 1)^n

    Correct

    (x^2 + (2x - 1))^n = x^{2n} + n(x^2)^{n-1}(2x - 1) + …\dots = x^{2n} + 2n x^{2n-1} + O\mathcal{O}(x^{2n-2})

    Exponents do not distribute over addition. For n≥2n \ge 2, (A+B)n≠An+Bn(A + B)^n \ne A^n + B^n. Expanding (x2+(2x−1))n(x^2 + (2x - 1))^n with the binomial theorem produces cross terms, the largest of which is (n1)(x2)n−1(2x−1)=2nx2n−1−nx2n−2\binom{n}{1}(x^2)^{n-1}(2x - 1) = 2n x^{2n-1} - n x^{2n-2}.

  3. Step 3

    g_n(x) = x^{2n} + (2x - 1)^n - x^{2n} = (2x - 1)^n   ⟹  \implies \deg(g_n(x)) = n

    Correct

    g_n(x) = [x^{2n} + 2n x^{2n-1} + O\mathcal{O}(x^{2n-2})] - x^{2n} = 2n x^{2n-1} + O\mathcal{O}(x^{2n-2})   ⟹  \implies \deg(g_n(x)) = 2n - 1

    Because the expansion in Step 2 omitted the 2nx2n−12n x^{2n-1} term, you calculated the remaining degree from (2x−1)n(2x - 1)^n, which incorrectly gave nn instead of 2n−12n - 1.

  4. Step 4

    lim⁡(deg⁡(gn(x)))=∞\lim (\deg(g_n(x))) = \infty

    Correct

    deg⁡(gn(x))=2n−1\deg(g_n(x)) = 2n - 1

    This final statement was carried over from the incorrect degree expression.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly recognized and applied the telescoping identity (x−1)(1+x+⋯+xn−1)=xn−1(x - 1)(1 + x + \dots + x^{n-1}) = x^n - 1.

You simplified fn(x)f_n(x) accurately by canceling the xnx^n terms to arrive at −1-1.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Illegal distribution of exponent over addition

no

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated (x2+(2x−1))n(x^2 + (2x - 1))^n as (x2)n+(2x−1)n(x^2)^n + (2x - 1)^n, applying power distribution across addition instead of using the binomial theorem to find the highest non-canceling cross term.

REASONING PATH
Your ACEpath teacher’s suggested approach
Binomial Expansion of Compound Bases

  • For any real or polynomial expressions AA and BB, (A+B)n=∑k=0n(nk)An−kBk(A + B)^n = \sum_{k=0}^n \binom{n}{k} A^{n-k} B^k, not An+BnA^n + B^n.

  • Group the polynomial by powers into A=x2A = x^2 and B=2x−1B = 2x - 1, then write out the k=0k = 0 and k=1k = 1 terms to capture the dominant remaining term after subtraction.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Group the terms as A+B=x2+(2x−1)A + B = x^2 + (2x - 1) and apply the binomial theorem: (x2+(2x−1))n=(n0)(x2)n+(n1)(x2)n−1(2x−1)+…(x^2 + (2x - 1))^n = \binom{n}{0}(x^2)^n + \binom{n}{1}(x^2)^{n-1}(2x - 1) + \dots

  2. Simplify the first two terms: x2n+nx2n−2(2x−1)=x2n+2nx2n−1−nx2n−2x^{2n} + n x^{2n-2}(2x - 1) = x^{2n} + 2n x^{2n-1} - n x^{2n-2}.

  3. Subtract x2nx^{2n} to find gn(x)g_n(x): gn(x)=2nx2n−1+…g_n(x) = 2n x^{2n-1} + \dots, where all subsequent terms have degree at most 2n−22n - 2.

  4. Since n≥1n \ge 1, the leading coefficient 2n≠02n \ne 0, so the highest power is 2n−12n - 1.

Final answer

fn(x)=−1f_n(x) = -1 for all n≥1n \ge 1, and deg⁡(gn(x))=2n−1\deg(g_n(x)) = 2n - 1.

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice expanding binomials (A+B)n(A + B)^n when AA has higher degree than BB, focusing on the first two terms: An+nAn−1BA^n + n A^{n-1} B.

  • Whenever the highest-degree term AnA^n cancels out in an expression, immediately check the next term (n1)An−1B\binom{n}{1} A^{n-1} B to identify the true leading degree.

Lock it in

Question

What does (x−1)(1+x+x2+⋯+xn−1)(x - 1)(1 + x + x^2 + \dots + x^{n-1}) expand to?

Answer

xn−1x^n - 1. Since the factor is (x−1)(x - 1), the leading term is x⋅xn−1=+xnx \cdot x^{n-1} = +x^n and the constant is −1⋅1=−1-1 \cdot 1 = -1.

Similar question

01

Simplify the expression P(x)=(x−1)(1+x+x2+x3+x4)−(x5−5)P(x) = (x - 1)(1 + x + x^2 + x^3 + x^4) - (x^5 - 5).

Answer

Since (x−1)(1+x+x2+x3+x4)=x5−1(x - 1)(1 + x + x^2 + x^3 + x^4) = x^5 - 1, we have P(x)=(x5−1)−(x5−5)=4P(x) = (x^5 - 1) - (x^5 - 5) = 4.

02

Simplify Q(x)=(x2−1)(1+x2+x4+⋯+x2n−2)−x2n+7Q(x) = (x^2 - 1)(1 + x^2 + x^4 + \dots + x^{2n-2}) - x^{2n} + 7.

Answer

Let u=x2u = x^2. The product becomes (u−1)(1+u+u2+⋯+un−1)=un−1=x2n−1(u - 1)(1 + u + u^2 + \dots + u^{n-1}) = u^n - 1 = x^{2n} - 1. Thus, Q(x)=(x2n−1)−x2n+7=6Q(x) = (x^{2n} - 1) - x^{2n} + 7 = 6.

03

Determine the degree and the leading coefficient of hn(x)=(x3+3x2−2)n−x3nh_n(x) = (x^3 + 3x^2 - 2)^n - x^{3n} for an integer n≥1n \ge 1.

Answer

By the Binomial Theorem, (x3+(3x2−2))n=x3n+n(x3)n−1(3x2−2)+O(x3n−2)=x3n+3nx3n−1+…(x^3 + (3x^2 - 2))^n = x^{3n} + n(x^3)^{n-1}(3x^2 - 2) + \mathcal{O}(x^{3n-2}) = x^{3n} + 3nx^{3n-1} + \dots. Subtracting x3nx^{3n} leaves the leading term 3nx3n−13nx^{3n-1}. Therefore, the degree is 3n−13n - 1 and the leading coefficient is 3n3n.

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