Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.6 - The Binomial Theorem - 12.6 Exercises - Page 886: 35

Answer

$ 300a^{2}b^{23}$

Work Step by Step

$(a+b)^{n}=\left(\begin{array}{l} n\\ 0 \end{array}\right)a^{n}+\left(\begin{array}{l} n\\ 1 \end{array}\right)a^{n-1}b+\left(\begin{array}{l} n\\ 2 \end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l} n\\ n \end{array}\right)b^{n}$ $\displaystyle \left(\begin{array}{l} n\\ r \end{array}\right)=\frac{n!}{r!(n-r)!},$ $n!=n(n-1)\cdot...\cdot 2\cdot 1,\ \ $ $n!=0$ Properties: $\left(\begin{array}{l} n\\ r \end{array}\right)=\left(\begin{array}{l} n\\ n-r \end{array}\right), $ $ \left(\begin{array}{l} k\\ r-1 \end{array}\right)+\left(\begin{array}{l} k\\ r \end{array}\right)=\left(\begin{array}{l} k+1\\ r \end{array}\right)$ -------- $...$In all the terms, exponents of a and b add to n... In $(a+b)^{25} $, which has 26 terms, the last three terms contain powers of a: 24th, 25th, 26th: $a^{2},\ \ \ a^{1},\ \ \ a^{0}$ The 24th term in the expansion of $(a+b)^{25}$ is where the exponent of $a$ is $2,$ the exponent of $b$ is $25-2=23,$ and the coefficient is $\left(\begin{array}{l} 25\\ 23 \end{array}\right)= \left(\begin{array}{l} 25\\ 25-23 \end{array}\right)= \displaystyle \left(\begin{array}{l} 25\\ 2 \end{array}\right)=\frac{25\cdot 24}{1\cdot 2}=300$ The term is: $ 300a^{2}b^{23}$
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