Find the constant term in the expansion of (2𝑥^6+−4/𝑥)^21. The constant term is ______

Question

More details. I have worked on this for over 1 hour I just can't get it plz help

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Answer 1

We will use the binomial expansion of (r + s)^t =

(1/0!) r^t s⁰ +

(1/1!) t r^(t-1) s¹ +

(1/2!) (t) (t-1) r^(t-2) s² +

(1/3!) (t) (t-1) (t-2) r^(t-3) s³ + . . .

where r = 2x⁶ and s = -4/x and t = 21

First we need to find where the constant term is. That will be where the power of x will be zero. Let that be term p, where p is an integer in the range of [0,21]. That term will look like

(1/p!) (t)(t-1)(t-2)...(t-p+1) (r)^(21-p) (s)^p

Letting r = 2x⁶ and s = -4/x, we get the p^th term:

(1/p!) (t)(t-1)(t-2)...(t-p+1) (2x⁶)^(21-p) (-4/x)^p

When 6(21-p) = p, the exponent of x will be reduce to zero. This happens when 6*21 = p+6p = 7p, or when p=18. Hence the constant term will be

(1/18!) (21)(20)(19)...(4) (2)^(21-18) (-4)¹⁸

or

(1/18!)(21!/3!)(2³⁹) = 731,175,232,471,040

Note: If you reverse r and s in this problem, p would be 3, and it might be easier to solve, but you'd get the same big answer.

Answer 2

I respectfully disagree with Mark M., who is usually right!!!!

The term with (2x⁶)³ * (4/x)¹⁸ will be a constant and that constant will be

-(21*20*19/6) * 8 * 2³⁶

That will be a darn big number (in absolute value) and I leave it to you to decide whether to actually try to write it or not.