Kuns742
Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Find the fifth term of (2x + y)^10

Sagot :

konrad509
[tex]T_r={n \choose r-1}a^{n-(r-1)}b^{r-1}\\T_5={10 \choose 5-1}(2x)^{10-(5-1)}y^{5-1}\\T_5={10 \choose 4}(2x)^6y^{4}\\T_5=\frac{10!}{4!6!}64x^6y^{4}\\T_5=\frac{7\cdot8\cdot9\cdot10}{2\cdot3\cdot4}64x^6y^{4}\\T_5=7\cdot3\cdot10\cdot64x^6y^{4}\\T_5=13440x^6y^4[/tex]
Everest2017
5th term in the expansion is

C(10,4)  (2x)^(10-4)  (y)^4

= 10C4  (2x)^6  y^4
=  10! / (6! 4! )    2^6  x^6 y^4
=  7 * 8 * 9 * 10 / 24    * 64  *x^6 y^4
= 13440 x^6  y^4


Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.