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Chapter 9: Problem 121

For \(x \in R, x \neq-1\), if \((1+x)^{2016}+x(1+x)^{2015}+x^{2}\) \((1+x)^{2014}+\ldots .+x^{2016}=\sum_{i=0}^{2016} a_{i} x^{i}\), then \(a_{17}\) is equal to : [Online April 9, 2016] (a) \(\frac{2017 !}{17 ! 2000 !}\) (b) \(\frac{2016 !}{17 ! 1999 !}\) (c) \(\frac{2016 !}{16 !}\) (d) \(\frac{2017 !}{2000 !}\)

Short Answer

Expert verified
The coefficient \(a_{17}\) is \(\frac{2017!}{17! \times 2000!}\); therefore, option (a).

Step by step solution

01

Analyze the Given Expression

The expression given is \((1+x)^{2016} + x(1+x)^{2015} + x^{2}(1+x)^{2014} + \ \ldots + x^{2016}\). We need to identify the pattern or series expansion to express it as a polynomial \(\sum_{i=0}^{2016} a_{i} x^{i}\), focusing on finding the coefficient \(a_{17}\).
02

Identify the Structure of the Expression

Consider the expression \((1+x)^{2016} + x(1+x)^{2015} + x^2(1+x)^{2014} + \ldots + x^{2016}\) as terms in a modified binomial expansion where each term can independently contribute terms to the power series.
03

Factor and Simplify the Expression

Observe that the expression can be rewritten as \((1+x)^{2016}\left[1+x+x^{2}+\ldots+x^{2016}\right]\). Here, the geometric series inside the brackets simplifies to \(\frac{1-x^{2017}}{1-x}\). Thus, the expression becomes \((1+x)^{2016} \cdot \frac{1-x^{2017}}{1-x}\).
04

Derive the Coefficient of \(x^{17}\)

The term \((1+x)^{2016}\) is a binomial expansion, where the coefficient of \(x^k\) is \(\binom{2016}{k}\). Using the simplified expression \((1+x)^{2016}\cdot (1-x^{2017})(1-x)^{-1}\), recognize that the coefficient of \(x^{17}\) involves the convolution of coefficients from \((1+x)^{2016}\) and \((1-x)^{-1}\).
05

Perform the Convolution

Recognize that the convolution needs calculations where \(x^{17}\) from \((1+x)^{2016}\) occurs with powers from \((1-x)^{-1}\). Specifically, the terms are summed as \(\binom{2016}{17} + \binom{2016}{16} + \ldots + \binom{2016}{0}\).
06

Use Binomial Sum Identity

Using the identity for partial sums of binomial coefficients, we recognize that the sum \(\binom{2016}{0} + \binom{2016}{1} + \ldots + \binom{2016}{17}\) equals \(\binom{2017}{17}\).
07

Calculate and Verify the Result

Find \(\binom{2017}{17} = \frac{2017!}{17! \, 2000!}\). Therefore, \(a_{17} = \frac{2017!}{17! \times 2000!}\), matching option (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Expansion
The concept of polynomial expansion involves breaking down a large polynomial into its constituent terms. This process is used to express powers of binomials, like \((1+x)^{2016}\), in simpler, expandable forms that are easier to analyze or compute.
  • In our expression, each term in the polynomial corresponds to a power series, where each power of \(x\) is multiplied by a specific coefficient.
  • By expanding these powers, we achieve a clearer form that simplifies calculations and helps in identifying specific coefficients, such as \(a_{17}\).
The expansion process will reveal correlations among different terms, based on their powers. It's fundamental to understanding how the binomial theorem applies to that polynomial.
Geometric Series
A geometric series is a series of terms each of which is obtained by multiplying the previous one by a fixed, non-zero number called the ratio. In the context of our polynomial expression, \(1+x+x^2+\ldots+x^{2016}\) represents a geometric series.
  • The formula for summing a geometric series is quite useful here and is expressed as \(\frac{1-r^{n+1}}{1-r}\), where \(r\) is the common ratio.
  • In this exercise, the sum can be written as \(\frac{1-x^{2017}}{1-x}\), exploiting the geometric series formula to simplify the entire polynomial expression.
This simplification is crucial for managing large polynomials because it transforms them into a more manageable form, allowing us to easily identify the necessary coefficients.
Binomial Coefficients
Binomial coefficients, denoted as \(\binom{n}{k}\), reflect the coefficients of terms in the expansions of binomials raised to powers, such as \((1+x)^{n}\).
These coefficients can be directly calculated using the formula:\[\binom{n}{k} = \frac{n!}{k! \,(n-k)!}\]
  • The relevance of binomial coefficients in our exercise is evident, as we are looking to compute the coefficient \(a_{17}\) by identifying relevant binomial coefficients.
  • The coefficients are essentially a way of determining the weight given to each term \(x^k\) in the expansion process.
Recognizing the role of binomial coefficients allows us to work systematically through complex expansions, simplifying computations by using well-established mathematical identities and properties. For instance, partial sums of binomial coefficients help derive results like \(\binom{2017}{17}\).This forms a solid foundation for solving the expansion problem.

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