Find the constant term in the expansion

Question

Find the constant term in the expansion of (-2x⁶+(5/x))²¹

what is the constant term?

Answer 1

(-2x⁶+(5/x))²¹=(-2x⁷+5)²¹/x²¹

The coefficient of x²¹ in the expansion of the numerator is what we want

(-2x⁷+5)²¹=∑_k=0^k=21(₂₁C_k)(-2x⁷)^k5^21-k

The term of interest is that for k=3 or ₂₁C₃(-8)5¹⁸=(-8)(21×20×19/(1×2×3))×5¹⁸

Excel gives me -40588378906250000 or -4.058837890625E+16

Answer 2

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)