Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.5 - The Binomial Theorum - Exercise Set - Page 1092: 43



Work Step by Step

Calculate the sixth term for $(x^2+y^3)^{8}$ by using General formula such as:$(m+n)^r=\displaystyle \binom{r}{k}m^{r-k}n^k$ and $\displaystyle \binom{r}{k}=\dfrac{r!}{k!(r-k)!}$ This implies,$(x^2+y^3)^{8}=\displaystyle \binom{8}{5}(x^2)^{8-5}(y^3)^5$ or, $=\dfrac{8!}{5!(8-5)!})x^{6}(y)^{15}$ or, $=[\dfrac{ 8 \times 7 \times 6 \times 5!}{5 !(3 \times 2 \times 1)}] x^{6}y^{15}$ or, $=56x^6y^{15}$
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