Sequences, Series and the Binomial Theorem
What this domain tests
Arithmetic and geometric progressions, limits of sequences, sums of finite and infinite series, and the binomial expansion with extraction of a specified term or coefficient.
Progressions and limits
- Arithmetic: a_n = a₁ + (n − 1)d and S_n = (n/2)(a₁ + a_n) = (n/2)(2a₁ + (n − 1)d).
- Geometric: a_n = a₁r^(n−1), S_n = a₁(r^n − 1)/(r − 1) for r ≠ 1, and S_∞ = a₁/(1 − r) only when |r| < 1.
- Limits: r^n → 0 when |r| < 1; r^n → 1 when r = 1; for rational expressions in n divide by the highest power and keep the leading coefficients. Also (1 + 1/n)^n → e.
- A monotone bounded sequence converges. Careful: a_n → 0 is necessary for a series to converge but not sufficient.
Binomial theorem
(a + b)^n = Σ from k = 0 to n of C(n,k)·a^(n−k)·b^k, where C(n,k) = n!/(k!(n−k)!).
- The general term is T_(k+1) = C(n,k)·a^(n−k)·b^k, with k counted from 0.
- C(n,k) = C(n, n−k) and Σ C(n,k) = 2^n.
- To find a coefficient, write the general term, collect the powers of x, set the exponent equal to the required value, solve for k, then substitute.
Worked example
Find the constant term in the expansion of (2x − 3/x²)^6.
General term: C(6,k)(2x)^(6−k)(−3/x²)^k = C(6,k)·2^(6−k)·(−3)^k·x^(6−k−2k) = C(6,k)·2^(6−k)·(−3)^k·x^(6−3k).
The constant term needs 6 − 3k = 0, so k = 2. The coefficient is C(6,2)·2⁴·(−3)² = 15·16·9 = 2160.
Traps that cost marks
- Using S_∞ when |r| ≥ 1, where the series diverges.
- Off-by-one in k: the exponent matching gives k, but the term is number k + 1.
- Losing the sign of (−3)^k when k is odd.
- Confusing "term" with "coefficient" when the question asks for one of them.
- Treating the binomial coefficients as symmetric in the powers of a and b incorrectly, i.e. swapping a and b without adjusting exponents.
- Assuming convergence from the mere fact that the individual terms shrink.