Kuns742
Kuns742 Kuns742
  • 06-08-2014
  • Mathematics
contestada

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

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konrad509
konrad509 konrad509
  • 06-08-2014
[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]
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Everest2017
Everest2017 Everest2017
  • 06-08-2014
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


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