Kuns742
Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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


Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.