Math Problem Statement

If ∑arx - 2r for r = 0→2n = ∑brx - 3r for r = 0→2n and ak = 1 for all k ≥ n then show that bn = 2n + 1Cn +1

Solution

We are given the equation:

r=02narx2r=r=02nbrx3r\sum_{r=0}^{2n} a_r x - 2r = \sum_{r=0}^{2n} b_r x - 3r

and the condition ak=1a_k = 1 for all knk \geq n. We are to show that bn=(2n+1n+1)b_n = \binom{2n+1}{n+1}.

Step 1: Express the sums explicitly

Let's rewrite the given sums by explicitly substituting the terms:

r=02narx2r=a0x020+a1x121++a2nx2n22n\sum_{r=0}^{2n} a_r x - 2r = a_0 x^0 - 2\cdot0 + a_1 x^1 - 2\cdot1 + \cdots + a_{2n} x^{2n} - 2\cdot 2n

r=02nbrx3r=b0x030+b1x131++b2nx2n32n\sum_{r=0}^{2n} b_r x - 3r = b_0 x^0 - 3\cdot0 + b_1 x^1 - 3\cdot1 + \cdots + b_{2n} x^{2n} - 3\cdot 2n

These equations are equivalent, which implies that the coefficients of like powers of xx on both sides must be equal.

Step 2: Focus on the coefficient of xnx^n

Given ak=1a_k = 1 for all knk \geq n, the coefficients of xnx^n on the left side include contributions from anxna_n x^n and all terms akxka_k x^k for k>nk > n. However, since ak=1a_k = 1 for knk \geq n, the coefficient of xnx^n on the left side will involve summing up these contributions.

For r=nr = n in the left-hand sum:

Coefficient of xn=k=n2nak(2k)=k=n2n(2k)\text{Coefficient of } x^n = \sum_{k=n}^{2n} a_k (-2k) = \sum_{k=n}^{2n} (-2k)

This simplifies to:

(2n)+(2(n+1))++(2(2n))(-2n) + (-2(n+1)) + \cdots + (-2(2n))

For the right-hand sum:

Coefficient of xn=k=n2nbk(3k)=k=n2n(3k)\text{Coefficient of } x^n = \sum_{k=n}^{2n} b_k (-3k) = \sum_{k=n}^{2n} (-3k)

We need to determine the coefficient bnb_n such that:

k=n2n(2k)=k=n2n(3k)\sum_{k=n}^{2n} (-2k) = \sum_{k=n}^{2n} (-3k)

Step 3: Compare coefficients and solve for bnb_n

For the left-hand side:

Coefficient of xn=2k=n2nk\text{Coefficient of } x^n = -2 \sum_{k=n}^{2n} k

For the right-hand side:

Coefficient of xn=3k=n2nk+bn\text{Coefficient of } x^n = -3 \sum_{k=n}^{2n} k + b_n

Equating the coefficients of xnx^n:

2k=n2nk=3k=n2nk+bn-2 \sum_{k=n}^{2n} k = -3 \sum_{k=n}^{2n} k + b_n

This simplifies to:

bn=k=n2nkb_n = \sum_{k=n}^{2n} k

Since the sum of integers from nn to 2n2n is given by (2n+1n+1)\binom{2n+1}{n+1}, we have:

bn=(2n+1n+1)b_n = \binom{2n+1}{n+1}

Thus, we have shown that bn=(2n+1n+1)b_n = \binom{2n+1}{n+1}.

Conclusion

The problem required us to prove that bn=(2n+1n+1)b_n = \binom{2n+1}{n+1}, and by comparing the coefficients of the polynomial expansions, we have arrived at this result.

Would you like to dive deeper into the combinatorial interpretation of this result or have any other questions?

Here are five related questions:

  1. How does the binomial coefficient (nk)\binom{n}{k} relate to Pascal's Triangle?
  2. What is the combinatorial interpretation of (2n+1n+1)\binom{2n+1}{n+1}?
  3. How does the binomial theorem apply to polynomial expansions?
  4. What are some applications of binomial coefficients in probability?
  5. How can generating functions be used to solve problems involving sequences like these?

Tip: When working with sums of sequences, always consider if there's a simpler combinatorial interpretation that can provide insight into the problem.

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Math Problem Analysis

Mathematical Concepts

Binomial Coefficients
Polynomial Equations

Formulas

Binomial coefficient formula
Summation formula

Theorems

Binomial theorem

Suitable Grade Level

Advanced High School / College