Most questions about an expansion want one term of it: the term in x⁵, the constant term, the middle term. The general term C(n,k)aⁿ⁻ᵏbᵏ answers them by solving one small equation for k. When that equation has no whole-number solution in range, the honest answer is that the term does not exist.
Core ideas
One expression and one equation. The difficulty is entirely in the bookkeeping, and the bookkeeping has three places to go wrong.
Choose k and the term is settled: the coefficient is C(n,k), the power of a is n - k, the power of b is k, and if b is negative the sign is (-1)ᵏ. Every question in this lesson is the question of which k is wanted.
When a and b carry powers of x, both contribute to the power of x in the term. In (2x² - 3/x)⁹ the a part contributes 2(9 - k) and the b part contributes -k, so the term holds x¹⁸⁻³ᵏ. Asking for a particular power of x is asking for the k that makes 18 - 3k equal to it.
The equation is exact and k counts brackets, so k = 16/3 is not a near miss to be rounded. It says the expansion contains no such term at all, which is a complete answer to the question that was asked.
The material
The general term, the exponent equation it leads to, and the cases where the term asked for is not there.
In the expansion of (a + b)ⁿ, the general term is C(n,k)aⁿ⁻ᵏbᵏ, where k runs from 0 to n. Written in descending powers of a, it is the (k + 1)-th term of the expansion.
The k + 1 is not decoration. k is a count of bs and starts at 0; the terms of a written-out expansion are counted from 1. Some books index the general term by the term number instead, putting r - 1 everywhere this course puts k. Both conventions are in use, and the safest habit is to describe a term by the power it contains rather than by its position.
If a and b are more than single letters, each contributes to the power of x in the term, and the two contributions have to be collected before anything can be solved. Take (x + 2/x)⁶, small enough to expand completely and check.
The power of x is 6 - 2k, which is 0 when k = 3. The term is C(6,3)·2³ = 20·8 = 160, and the table above shows the same 160 in the row for k = 3. Six of the seven terms never had to be computed to answer the question.
The equation to solve is always the same shape. Collect the power of x in the general term, set it equal to the power wanted, and solve. The exponent is linear in k, and unless a and b carry the same power of x — in which case every term carries it too, and there is nothing to solve — there is exactly one candidate. It then has to survive two tests: it must be a whole number, and it must lie between 0 and n.
In (x² + 1/x)⁸ the general term is C(8,k)x¹⁶⁻²ᵏx⁻ᵏ = C(8,k)x¹⁶⁻³ᵏ. Setting 16 - 3k = 0 gives k = 16/3, which is not a whole number. The powers of x in the nine terms run 16, 13, 10, 7, 4, 1, -2, -5, -8, stepping over 0 between the sixth and seventh terms. The expansion has no constant term, and that is the answer.
An expansion of (a + b)ⁿ has n + 1 terms. If n is even there is a single middle term, at k = n/2. If n is odd there are two, at k = (n - 1)/2 and k = (n + 1)/2, and their binomial coefficients are equal by the symmetry of the triangle — but the terms themselves need not be, once a and b carry numbers.
n = 7, so the middle terms are k = 3 and k = 4, and C(7,3) = C(7,4) = 35. At k = 3: 35(2x)⁴(-1)³ = 35·16·(-1)x⁴ = -560x⁴. At k = 4: 35(2x)³(-1)⁴ = 35·8x³ = 280x³. Equal binomial coefficients, and terms differing in size and in sign: the symmetry belongs to C(n,k), and (2x)⁴ against (2x)³, (-1)³ against (-1)⁴, breaks it.
Signs come from b and nowhere else. Writing b = -3 rather than carrying a minus sign outside means the factor (-3)ᵏ handles the alternation with no separate rule to remember, and an even k returns a positive contribution however negative b is.
One caution about checking. Substituting x = 1 turns the whole expansion into the sum of its coefficients, and comparing that with the bracket evaluated at x = 1 is a strong check on a complete expansion. It says nothing about a single term computed on its own: if you have written down one term out of ten, there is nothing for the substitution to check against. What checks a single term is recomputing its exponent from the two factors separately.
Interactive
C(n, k) a^(n-k) b^k on its own, and then the expansion that places it
Method
Four steps. The third can end the question with the answer that no such term exists.
In (2x² - 3/x)⁹: a = 2x², b = -3/x, n = 9. The minus belongs to b. Writing this line down before anything else is what keeps the sign out of the coefficient.
C(n,k)aⁿ⁻ᵏbᵏ, then take the numbers and the powers of x apart. Both a and b contribute to the exponent, and the total is a linear expression in k.
Then check the solution twice: it must be a whole number, and it must lie in 0 to n. If it fails either test, the expansion has no such term and the question is answered.
The binomial coefficient, the power of the numerical part of a, and the power of b including its sign. Multiplying them in that order keeps the arithmetic small and makes a wrong sign visible.
A constant term is the term whose power of x is 0. Nothing else in this expansion has to be computed, and the ten terms it contains would take a page.
The exponent check is the one that matters, because it recomputes the power of x from the two factors separately rather than reusing the expression 18 - 3k that produced k = 6. A slip in collecting the exponent would survive the first calculation and fail this one.
Checking the size of the number would not have helped here. The k = 5 term of the same expansion is -489888x³: the same magnitude with the opposite sign, because C(9,5)·2⁴·3⁵ = 126·16·243 is also 489888. Two adjacent terms happening to agree in size is a coincidence of this expansion, and it is a good reason to pin k by the exponent rather than by recognising a number.
For a faded single-term pass, find the constant term of (x³ - 2/x)⁸. The supplied identification is a = x³, b = -2/x, n = 8. Write the general term, collect both contributions to the power of x, solve the resulting equation for k, and check that k is a whole number in range. Recompute the final exponent from the two powered factors separately before opening the quiz.
You should be able to write the general term of any bracket, collect its power of x from both parts, solve for k, and say plainly when no such term exists. Getting C(n,k) from a row of the triangle or from the factorial formula, and the sign from (-1)ᵏ, should be routine by now.
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