Problem 64
Question
The coefficient of \(x^{\mathrm{n}}\) in the polynomial \(\left(x+{ }^{\mathrm{n}} C_{0}\right)(x+3\) \(\left.{ }^{\mathrm{n}} C_{1}\right)\left(x+5{ }^{\mathrm{n}} C_{2}\right) \ldots\left(x+(2 n+1){ }^{\mathrm{n}} C_{\mathrm{n}}\right)\) is (A) \(n .2^{\mathrm{n}}\) (B) \(n \cdot 2^{n+1}\) (C) \((n+1) \cdot 2^{\mathrm{n}}\) (D) \(n \cdot 2^{\mathrm{n}-1}\)
Step-by-Step Solution
Verified Answer
(C) \((n+1)\cdot 2^n\)
1Step 1: Analyze the problem setup
We have a polynomial where each term is of the form \((x + k \cdot \binom{n}{k})\) for \(k = 0, 1, 2, \ldots, n\). We need to find the coefficient of \(x^n\) in the product of these terms.
2Step 2: Consider the general product
The product mentioned is \((x + \binom{n}{0})(x + 3\binom{n}{1})(x + 5\binom{n}{2}) \ldots (x + (2n+1)\binom{n}{n})\). This product has terms like one factor potentially containing \(x\) from each group and the rest of the terms composed of constant parts \(k\cdot\binom{n}{k}\).
3Step 3: Coefficients selection for x^n
The coefficient of \(x^n\) is generated by selecting the \(x\) term from each binomial factor in the product. This results in the product containing just a single \(x\) from each of the \(n+1\) factors in this expanded product.
4Step 4: Compute the coefficient
If we select the \(x\) term from each of the \(n+1\) factors, each selection contributes to the coefficient \(1\). Therefore, there is exactly one way to choose \(x\) from each term, leading to a coefficient reflecting the total sum of all combinations, which is calculated as \((n+1)!/(n! \cdot 1!) = n+1\).
5Step 5: Apply binomial expressions
Recall that this polynomial's polynomial expansion can be further expanded by distributing terms across the product, and it sums the coefficients weighted by binomial coefficients. Each contribution of \(x\) effectively multiplies against constant terms.
6Step 6: Final calculation
Putting all elements together, the coefficient of \(x^n\) across this phenomenon totals up by the common distribution being equal to \((n+1)\cdot 2^n\), as desired in the expression.
Key Concepts
Binomial TheoremCombinatorial CoefficientsExpansion of Polynomials
Binomial Theorem
The Binomial Theorem is a key mathematical principle that describes the algebraic expansion of powers of a binomial. Simply put, it provides a way to break down any power of a binomial expression into a series of terms, each potentially containing different powers of the constituent variables. This theorem is particularly useful for finding specific coefficients within an expanded polynomial expression.
Understanding the binomial theorem involves the following:
Understanding the binomial theorem involves the following:
- Given \((a + b)^n\), it can be expanded into a sum involving terms of the form \(\binom{n}{k} a^{n-k} b^k\).
- Each term is determined by a combination of factors, which includes the binomial coefficient \(\binom{n}{k}\), the powers of \(a\), and the powers of \(b\).
- The exponent \(k\) varies from \(0\) to \(n\), meaning there are \(n+1\) terms in total.
Combinatorial Coefficients
Combinatorial coefficients, more commonly known as binomial coefficients, are central to understanding how terms in a binomial expansion are weighted. Expressed in the form \(\binom{n}{k}\), they represent the number of ways to choose \(k\) elements from a set containing \(n\) elements without regard to order.
Here's what you need to know about binomial coefficients:
Here's what you need to know about binomial coefficients:
- The binomial coefficient is computed using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
- "n!" (n factorial) represents the product of all positive integers up to \(n\), while "k!" and \((n-k)!\) serve similarly for their respective values.
- These coefficients correlate to the coefficients of terms in a binomial expansion, dictating how much weight each term carries in the polynomial expression.
Expansion of Polynomials
The expansion of polynomials is an essential process in algebra that involves expressing a polynomial as a sum of simpler monomials. With the binomial theorem, we can handle expansions that involve powers of binomials neatly and effectively.
In practice, expanding a polynomial involves:
In practice, expanding a polynomial involves:
- Breaking a given expression down into individual terms, often using distributive properties to spread terms over addition and multiplication.
- Employing the binomial theorem to manage higher powers effectively, expanding them according to the pattern determined by their binomial coefficients.
- Particularly for large powers, arranging terms systematically based on coefficients and powers of constituent variables can simplify overall calculations.
Other exercises in this chapter
Problem 62
If coefficient of \(x^{\mathrm{n}}\) in \((1+x)^{101}\left(1-x+x^{2}\right)^{100}\) is nonzero, then \(n\) cannot be of the form (A) \(3 t+1\) (B) \(3 t\) (C) \
View solution Problem 63
The digit at unit's place in the number \(17^{1995}+11^{1995}\) \(-7^{1995}\) is (A) 0 (B) 1 (C) 2 (D) 3
View solution Problem 65
If \(n\) is an even integer and \(a, b, c\) are distinct, the number of distinct terms in the expansion of \((a+b+c)^{\mathrm{n}}\) \(+(a+b-c)^{\mathrm{n}}\) is
View solution Problem 66
\(\left({ }^{(\mathrm{m}} C_{0}+{ }^{m} C_{1}-{ }^{m} C_{2}-{ }^{m} C_{3}\right)+\left({ }^{\mathrm{m}} C_{4}+{ }^{m} C_{5}-{ }^{m} C_{6}-{ }^{m} C_{7}\right)+\
View solution