/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Find the coefficient of \(x^{20}... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 7

Find the coefficient of \(x^{20}\) in \(\left(x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right)^{5}\).

Short Answer

Expert verified
The coefficient of \(x^{20}\) in the given polynomial raised to the fifth power is 26.

Step by step solution

01

Understanding the Binomial Theorem

The binomial theorem explains how to expand a binomial raised to any power. In essence, it gives us ratios for each term in the expanded binomial. This theorem can be extended to polynomials with more terms.
02

Expanding the Polynomial

The polynomial \(\left(x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right)^{5}\) is raised to the power of five. This means each term from the original polynomial may appear up to five times. That creates combinations of situations, like having five \(x^2\)鈥檚 or three \(x^2\)鈥檚 and two \(x^4\)鈥檚, and so on.
03

Finding the Coefficient of \(x^{20}\)

Now look for combinations of terms whose powers add up to 20. These are (1) five \(x^4\)鈥檚, (2) one \(x^2\), one \(x^3\) and three \(x^5\)鈥檚 and (3) two \(x^2\)鈥檚 and three \(x^6\)鈥檚. Considering the combinatorial coefficients, the coefficient of \(x^{20}\) is given by \(C(5,5)+C(5,1)*C(4,1)*C(3,3) + C(5,2)*C(3,3)\)
04

Computing the Result

By substituting the appropriate factorials in the combinatorial expressions, the coefficients are calculated as \(1 + (5*4*1) + (10*1) = 26\).

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Most popular questions from this chapter

Suppose that \(Y\) is a geometric random variable where the probability of success for each Bernoulli trial is \(p\). If \(m, n \in \mathbf{Z}^{+}\) with \(m>n\), determine \(\operatorname{Pr}(Y \geq m \mid Y \geq n)\).

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