Find the constant term in the expansion of (-2x^6+(5/x))^21.
Can't divide by 0 ==> x ≠ 0.
f(x) = (-2x^6+(5/x))^21 =
+ C(21,0) (-2x^6)^21 (5/x)^0
+ C(21,1) (-2x^6)^20 (5/x)^1
+ C(21,2) (-2x^6)^19 (5/x)^2
+ C(21,3) (-2x^6)^18 (5/x)^3
+ C(21,4) (-2x^6)^17 (5/x)^4
...
+ C(21,k) (-2x^6)^(21-k) (5/x)^k
The constant term has a 0 exponent for the x factor, so:
x^(6(21-k)-k) = x^0
6(21-k)-k = 0
6(21) – 6k – k = 0
7k = 6(21)
k = 18
So the constant term is:
C(21,18) (-2x^6)^(21-18) (5/x)^18
= C(21,18) (-8x^18) (5^18/x^18)
= (-8)(5^18) C(21,18)
= (-8)(5^18)(21!/(18!(21-18)!))
= (-8)(5^18)(21*20*19*18!/(18!(3)!))
= (-8)(5^18)(21*20*19/(3*2))
= –4(5^18)(7*20*19)
= –4(3814697265625)(2660)
≈ –4.058837890625(10^16)
Michael F.
(-2x6)3=-8x18
04/03/14