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(a) Without expanding find the constant term in the expansion of (4x-(1/3)x²)^18.
(b) Without expanding find the term involving x^6 in the expansion of (4x-(1/3)x²)^18.
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(a) The general formula for the k-th term of this expansion is
= , k = 0, 1, 2, 3, . . . , 18.
The constant term is the term with (18-k) + 2k = 0, or 18 + k = 0, k = -18.
But this value, k= -18 is not in the range [0,18] of possible values of 'k'.
So, we conclude that the constant term absents in this expansion.
It is the same as to say that the constant term has the coefficient '0' in the expansion.
It becomes obvious as you look at the basic binomial: you will see that x does present
in both terms of the binomial; so, the constant term, independent of x, absents in the expansion.
(b) Again, the general formula for the k-th term of this expansion is
= , k = 0, 1, 2, 3, . . . , 18.
The term involving x^6 is the term with (18-k) + 2k = 6, or 18 + k = 6, k = 6 - 18 = -12.
But this value, k= -12 is not in the range [0,18] of possible values of 'k'.
So, we conclude that the constant term absents in this expansion.
It is the same as to say that the term involving x^6 has the coefficient '0' in the expansion.
It becomes obvious as you look at the basic binomial: you will see that x does present
in both terms of the binomial; so, the degrees of 'x' in the expansion all are at least 18.
Solved.