General and middle terms

What is the general term?

In plain words: you do not need the whole expansion to find one term. The term with b^r comes from choosing which r of the n brackets give b, so its coefficient is the number of combinations ⁿC_r. Because the first term has r = 0, the term number is always one more than r: the 4th term has r = 3.

When the second part is negative, keep the minus sign inside b. For (x − y)ⁿ, take b = −y, so the sign of each term is (−1)^r:

The kth term from the end of (a + b)ⁿ is the (n − k + 2)th term from the start, since there are n + 1 terms in all. It is also the kth term of (b + a)ⁿ.

The middle term or terms

(a + b)ⁿ has n + 1 terms, so the number of middle terms depends on whether n is even or odd.

• n even: n + 1 is odd, and there is one middle term, the (n/2 + 1)th. For n = 8 it is the 5th term, T₅.
• n odd: n + 1 is even, and there are two middle terms, the ((n + 1)/2)th and the ((n + 3)/2)th. For n = 7 they are T₄ and T₅.

The middle binomial coefficients are the largest. ⁿC_r rises to the middle because each step multiplies it by (n − r + 1)/r, which is more than 1 while r < (n + 1)/2, and falls back symmetrically because ⁿC_r = ⁿC_n−r (see the properties of ^ⁿC_ᵣ). This is about ⁿC_r only: when a and b carry numbers, the greatest coefficient can be elsewhere. In (1 + 2x)⁷ the middle terms are T₄ and T₅, but the greatest coefficient is 672, in T₆.

Finding a given term or coefficient

To find the term with a given power of x, or the term independent of x (the constant term), use the general term:

1. Write T_r+1 and collect all the powers of x into one power x^{(something in r)}.
2. Set that power equal to the power you want (0 for the term independent of x) and solve for r.
3. r must be a whole number from 0 to n. If it is not, there is no such term.
4. Put r back into T_r+1. The coefficient is the number in front, including its sign.

Ratios of neighbouring terms are often quicker than the terms themselves. For 1 ≤ r ≤ n, use a term ratio only when its denominator T_r is non-zero (and a ≠ 0 in the expression below). In (a + b)ⁿ, T_r+1/T_r = ((n − r + 1)/r) × b/a; for (1 + x)ⁿ the coefficients are in the ratio (n − r + 1)/r:

Solved examples

Common mistakes to avoid

Quick revision summary

• The general term of (a + b)ⁿ is T_r+1 = ⁿC_ra^n−rb^r; the term number is r + 1.
• n even: one middle term, the (n/2 + 1)th; n odd: two, the ((n + 1)/2)th and ((n + 3)/2)th. Their binomial coefficients ⁿC_r are the largest; with numbers inside a or b, the greatest coefficient can be in another term.
• For a given power of x, solve for r in the general term; if r is not a whole number from 0 to n, there is no such term.

Questions students ask

How do you find the greatest term in a binomial expansion?

For a positive integer n and real x ≠ 0, the term of greatest absolute value in (1 + x)ⁿ, beyond NCERT, comes from the ratio of neighbouring terms: |T_r+1/T_r| ≥ 1 exactly when r ≤ m = (n + 1)|x|/(1 + |x|). If m is not a whole number, the greatest term is T_r+1 with r the whole-number part of m; if it is, T_m and T_m+1 have equal, greatest absolute values; for negative x their signs are opposite. When x = 0, the ratio method is unnecessary: T₁ = 1 and all later terms are 0. For (1 + x)¹⁰ at x = 1/2, m = 11/3, so the greatest term is T₄ = 15, not the middle term T₆ ≈ 7.9. For (a + bx)ⁿ with a ≠ 0, use |bx/a| in place of |x|, or |b/a| for the coefficient of greatest absolute value. If a = 0, the expression is simply (bx)ⁿ; if the ratio is 0, only the first formal term can be non-zero.

What does ‘term independent of x’ mean?

The term independent of x is the constant term of an expansion: the term in which the power of x is 0, so its value is the same for every x ≠ 0. In (x − 2/x)⁶ the general term is ⁶C_r(−2)^rx^6−2r, so r = 3 gives the term independent of x, ⁶C₃(−2)³ = −160. When no whole number r from 0 to n makes the power of x zero, the expansion has no term independent of x, which means its constant term is 0.

What is the middle term of (1 + x)^²ⁿ?

The middle term of (1 + x)²ⁿ is the (n + 1)th term, T_n+1 = ²ⁿC_nxⁿ, because the expansion has 2n + 1 terms and an even index gives a single middle term. Its coefficient ²ⁿC_n is the greatest binomial coefficient of the expansion, and it can also be written in the classic form [1 · 3 · 5 … (2n − 1)] × 2ⁿ/n!. For n = 3, the middle term of (1 + x)⁶ is ⁶C₃x³ = 20x³, and indeed 1 · 3 · 5 × 2³/3! = 120/6 = 20.

How is the general term of a binomial expansion different from that of an AP or GP?

The general term of a binomial expansion, T_r+1 = ⁿC_ra^n−rb^r, does the same job as the general term a + (n − 1)d of an AP or arⁿ⁻¹ of a GP: it gives the term in any position. The letters differ: in an AP or GP, n is the position, a the first term and r the common ratio, while in (a + b)ⁿ, n is the power, a one part of the binomial and r + 1 the term number. A binomial expansion also stops after n + 1 terms, while an AP or GP can be infinite.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the 5th term of (2x − y)⁶.

   Show answerHide answer

   r = 4: ⁶C₄(2x)²(−y)⁴ = 15 × 4x²y⁴ = 60x²y⁴.

2. Q2. Find the middle term of (x + 1/x)¹⁰.

   Show answerHide answer

   n = 10 is even, so the middle term is T₆ (r = 5): ¹⁰C₅x⁵(1/x)⁵ = 252.

3. Q3. Find the coefficient of x⁶ in (1 + 2x)⁹.

   Show answerHide answer

   T_r+1 = ⁹C_r(2x)^r, so r = 6: ⁹C₆ × 2⁶ = 84 × 64 = 5376.

4. Q4. Find the term independent of x in (2x + 1/x²)⁶.

   Show answerHide answer

   T_r+1 = ⁶C_r2^6−rx^6−3r. 6 − 3r = 0 gives r = 2: T₃ = 15 × 16 = 240.