What is the third term in the expansion of (x - 3)^7 ?
How do i solve this?
2007-12-06 03:01:56 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^7 [ 1 - (3/x) ]^(7)
x^7 [1 - 7 (3/x ) + ( (7) (6) / 2! ) 9/x² - --------]
Third term is 189 x^5
2007-12-06 03:28:18 · answer #1 · answered by Como 7 · 2⤊ 0⤋
Use the Binomial theorem: (7 C 2) (x^(7-2))(-3)^2
7 C 2 = 7! /(2! 5!) = 21
21(x^5)(9)
189x^5
2007-12-06 03:09:20 · answer #2 · answered by Linda K 5 · 0⤊ 1⤋
Using the binomial theorem. (x + y)^n is equal to
n
the sum of C(n,k) *x^(n-k)*y^k
k = 0
So in your case, C(7,2) (x^(7-2))(-3)^2
2007-12-06 03:11:34 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Use binomial theorem
2007-12-06 03:04:56 · answer #4 · answered by Kaaks 3 · 1⤊ 0⤋
7C2(x^5)(-3)^2 = 189x^5
2007-12-06 03:11:12 · answer #5 · answered by Anonymous · 0⤊ 0⤋