What is the 6th term in the expression (2x-y)^10

Hello from MrBillDoesMath!

- 8064 x^5 y^5

Discussion:

The binomial theorem gives us the expansion  of  (a + b) ^n  :

(a + b ) ^n

= a^n + na^(n-1)b +  (n(n-1)/ 2!) a^(n-2)b^2 +

In our case, taking the 6th term above, setting a = 2x, b = -y, and n = 10 gives

( (10*9*8*7*6)/ (5 *4*3*2*1) )  * (2x)^5 (-y)^5r

=  - 8064 x^5 y^5

Thank you,

MrB

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000