K5MA4 · Sequences series binomial expansion and models · Edexcel A-level Mathematics

Revise the key ideas

Sequence definitions recurrence and sigma notation

Sequence and index — A sequence is an ordered list of terms associated with an index, usually positive integers unless stated otherwise. If uₙ = 3n + 1 for n ≥ 1, the first terms are 4,7,10. Check whether a model starts at n = 0 or n = 1 before deciding which formula value is the initial observation.
Explicit rule — An explicit nth-term rule gives a term directly from its index. For uₙ = n² − n + 2, u₅ = 22 without calculating the four earlier terms. Different-looking expressions can define the same sequence, but a handful of matching terms alone does not prove two general rules identical.
Recurrence and initial value — A recurrence generates a new term from preceding information and needs enough initial data. With u₁ = 2 and uₙ₊₁ = 2uₙ − 1, the sequence begins 2,3,5,9. The formula alone allows many sequences; changing the starting value changes the resulting terms.
Increasing and decreasing — A sequence is strictly increasing if uₙ₊₁ > uₙ at every relevant index and strictly decreasing if the reverse holds. Nondecreasing allows equal adjacent terms. Compare a general difference uₙ₊₁ − uₙ or use a suitable argument; two early terms do not establish behaviour for all indices.
Difference test — For uₙ = 1/(3n + 1), subtract consecutive terms: uₙ₊₁ − uₙ = −3/((3n + 4)(3n + 1)) < 0 for n ≥ 1. The sequence therefore decreases and remains positive. Stating the denominator's sign is part of the justification.
Periodic sequence — A periodic sequence repeats after a fixed positive integer number of steps. With u₁ = 3 and uₙ₊₁ = 1/uₙ, the values alternate 3,1/3,3,1/3, so the smallest period is 2. A starting value of 1 instead gives a constant sequence of period 1; check the initial value and undefined cases.
Limit and oscillation — Convergence means terms approach a finite limit as the index grows. A sequence can converge while oscillating, such as uₙ = (−1)ⁿ/n approaching zero. A periodic sequence alternating 3 and 1/3 does not converge because the two subsequences approach different values. Bounded means all terms stay within fixed finite limits. For example, 3,1/3,3,1/3,… is bounded but does not converge, so the two properties are different.
Sigma meaning — The notation Σ from k = 1 to n of aₖ means a₁ + a₂ + … + aₙ. The index k is a dummy summation variable; n is an upper bound. Writing Σ from k = 1 to n of 1 gives n, while summing 1 from k = 0 to n gives n + 1.
Expanding a sum — The sum Σ from k = 2 to 5 of (2k + 1) is 5 + 7 + 9 + 11 = 32. Include both endpoints unless the notation says otherwise. The number of integer terms from k = p to q with q ≥ p is q − p + 1, not q − p.
Linearity and shifted bounds — Finite sums distribute over addition and fixed multipliers. Thus Σ from k = 1 to n of (3k + 2) equals 3Σk + 2n. A sum from k = p to q can be found by subtracting the sum to p − 1 from the sum to q, using the same summand. Take care with index shifts that alter the first term.

Arithmetic sequences series and proof

Common difference — An arithmetic sequence has a fixed difference d between successive terms. Its nth term is uₙ = a + (n − 1)d when the first term is a at n = 1. For 7,11,15,…, a = 7 and d = 4, so uₙ = 4n + 3. The n − 1 counts the number of steps after the first term.
Finding arithmetic parameters — If an arithmetic sequence has u₃ = 11 and u₈ = 31, then a + 2d = 11 and a + 7d = 31. Subtract to obtain d = 4 and a = 3. The gap between the indices is five, so divide the term difference twenty by five, not eight or three.
Term threshold — For uₙ = 7 + 4(n − 1), the first term exceeding 100 satisfies 4n + 3 > 100, so n > 97/4 = 24.25. The first integer index is 25 and u₂₅ = 103. If the question asks at least rather than strictly above, adjust the inequality before rounding.
Arithmetic sum formula — The sum of the first n arithmetic terms is Sₙ = n(2a + (n − 1)d)/2, equivalently n(a + l)/2 where l is the nth term. For a = 7, d = 4 and n = 20, the last term is 83 and S₂₀ = 20(7 + 83)/2 = 900. A sum and an nth term are different quantities.
Proof by reversal — Write Sₙ = a + (a + d) + … + (a + (n − 1)d), then write the same sum in reverse order. Adding corresponding terms gives 2Sₙ = n(2a + (n − 1)d), because every pair has the same value and there are n pairs. Dividing by two derives the formula rather than merely quoting it.
The same arithmetic sum is written forward and backward. Each vertical pair equals a+l, and there are n such pairs, so the doubled sum is n(a+l). The final term is l=a+(n−1)d, giving both standard sum forms. The ellipsis stands for the full set of intermediate terms rather than three terms only.
Natural-number sum — For 1 + 2 + … + n, put a = 1 and d = 1 in the arithmetic formula to obtain n(n + 1)/2. The reversal proof pairs the first with the last to give n + 1 for every column. This also explains why the sum is an integer even though the expression contains a division by two.
Partial arithmetic sum — To sum arithmetic terms from index 6 through 20, subtract S₅ from S₂₀, or treat u₆ as the first of fifteen terms. For uₙ = 4n + 3, u₆ = 27, u₂₀ = 83 and the total is 15(27 + 83)/2 = 825. Counting the endpoints avoids an off-by-one error.
Solving from a sum — If an arithmetic series has a = 3, d = 2 and Sₙ = 120, then Sₙ = n(n + 2), so n² + 2n − 120 = 0. Its roots are 10 and −12; only the positive integer n = 10 is an admissible number of terms. A real algebraic root is not automatically a usable count.
Decreasing arithmetic terms — A negative d produces decreasing terms, but the sum can still increase while the added terms remain positive. With a = 20 and d = −3, terms are 20,17,14,… . Distinguish monotonic behaviour, always moving in one direction or remaining equal, of uₙ from that of Sₙ; adding a negative later term decreases the partial sum.
Interpreting arithmetic growth — A constant additive change suits a model only where the same amount is added each step. Increasing by four units per month is arithmetic; increasing by four percent of the current amount is generally geometric. State the unit of the term and the time represented by one index step.

Geometric sequences finite sums and convergence

Common ratio — A geometric sequence multiplies each term by a fixed ratio r, giving uₙ = arⁿ⁻¹ for first term a at n = 1. For 6,3,3/2,…, a = 6 and r = 1/2. Finding a ratio by division requires the previous term to be nonzero; handle a zero starting sequence separately.
Finding a ratio — If u₂ = 12 and u₅ = 96 in a nonzero geometric sequence, dividing gives r³ = 8, so r = 2 and a = 6. If an even power of r is inferred instead, positive and negative ratio possibilities may both survive unless signs or context rule one out. Keep all real branches until checking the data.
Finite geometric sum — For r ≠ 1, Sₙ = a(1 − rⁿ)/(1 − r), equivalently a(rⁿ − 1)/(r − 1). With a = 6, r = 1/2 and n = 4, S₄ = 45/4 = 11.25. The value includes four terms 6 + 3 + 3/2 + 3/4.
Proof by subtraction — Write Sₙ = a + ar + … + arⁿ⁻¹ and multiply by r to get rSₙ = ar + ar² + … + arⁿ. Subtracting gives (1 − r)Sₙ = a − arⁿ, so division by 1 − r yields the finite formula when r ≠ 1. Show which intermediate terms cancel and retain the final power rⁿ.
For four terms, multiplying by r shifts every power up by one. Subtraction cancels ar, ar² and ar³, leaving a−ar⁴. With n terms the retained final power is rⁿ, and division by 1−r requires r not equal to one. The stacked continuation of the rS₄ row is part of the same expression.
Ratio one — For r = 1, every term equals a and Sₙ = na. The usual quotient formula gives a zero denominator and cannot be used by direct substitution. If a ≠ 0 the infinite series does not converge; if a = 0 the all-zero series has zero sum without needing the usual nonzero-series criterion.
Negative ratios — A negative ratio alternates term signs when a ≠ 0. With a = 8 and r = −1/2, the first four terms are 8,−4,2,−1 and their sum is 5. The finite formula remains valid; alternating signs do not imply every partial sum or term is decreasing.
Sum to infinity — For a nonzero geometric series with the modulus of r < 1, rⁿ tends to zero and S∞ = a/(1 − r). With a = 6 and r = 1/2, the sum is 12. The condition concerns convergence of the infinite sequence of partial sums, not a finite calculation involving a very large chosen n.
Convergence condition — Both positive ratios between zero and one and negative ratios between minus one and zero satisfy the modulus of r < 1. For r = −1 the terms alternate without shrinking and partial sums generally do not approach one limit; for r = 1 nonzero terms accumulate without a finite limit. Check the strict boundary rather than use the modulus of r ≤ 1.
Negative-ratio infinite sum — For a = 8 and r = −1/2, S∞ = 8/(1 + 1/2) = 16/3. A sum to infinity can be smaller than the first positive term because later negative terms subtract. It is the partial sums that approach 16/3, while the individual terms approach zero.
For first term 8 and ratio −1/2, the finite totals alternate around their limit 16/3. The individual terms instead approach zero. The displayed values are finite partial sums, not individual terms, illustrating why a term limit and a series sum must be distinguished.
Remainder after n terms — For a convergent geometric series, S∞ − Sₙ = arⁿ/(1 − r). Its magnitude estimates the exact remaining total, while its sign can alternate when r is negative. With a = 6 and r = 1/2, the remainder after n terms is 12(1/2)ⁿ; after ten terms it is 3/256.
Logarithmic term count — To make that positive remainder less than 0.01, solve 12(1/2)ⁿ < 0.01. Taking logarithms gives n > ln(0.01/12)/ln(1/2), approximately 10.23, with reversal when dividing by a negative logarithm. The first possible integer is 11; check ten and eleven in the original inequality.
Solving a geometric sum — If 3(1 − (1/2)ⁿ)/(1 − 1/2) = 189/32, then 6(1 − 2⁻ⁿ) = 189/32, so 2⁻ⁿ = 1/64 and n = 6. Rearrange to isolate the power before taking logs. Check that n is a nonnegative or positive integer as required by the indexing convention.

Binomial coefficients expansions and validity

Factorial and combinations — For a positive integer n, n! is the product 1×2×…×n, with 0! = 1. The binomial coefficient n choose r is n!/(r!(n − r)!) for integer 0 ≤ r ≤ n. It counts selections of r positions from n without ordering, which explains its role in expanding repeated factors.
Positive-integer expansion — For nonnegative integer n, (a + bx)ⁿ = Σ from r = 0 to n of (n choose r)aⁿ⁻ʳ(bx)ʳ. The first summand has r = 0 and the final one r = n. Powers of both a and b matter: do not insert only coefficients from Pascal's triangle and forget the scaling.
Pascal coefficients — Each interior binomial coefficient is the sum of the two above it in Pascal's triangle, and n choose r equals n choose n − r. For n = 5 the row is 1,5,10,10,5,1. This follows from choosing whether a fixed position is included in a selection, not a new multiplication rule for coefficients.
Worked finite expansion — For (2 − x)⁵, the full expansion is 32 − 80x + 80x² − 40x³ + 10x⁴ − x⁵. The rth chosen-position term is (5 choose r)2⁵⁻ʳ(−x)ʳ, so odd powers are negative. This is a finite identity valid for every real x, with no small-input restriction.
Coefficient versus term — In (2 − x)⁵, the coefficient of x³ is −40 while the x³ term is −40x³. The fourth term corresponds to r = 3 because counting begins at the constant term r = 0. State whether a question asks for a numerical coefficient, a term or a position.
General rational expansion — For rational p, (1 + u)ᵖ = 1 + pu + p(p − 1)u²/2! + p(p − 1)(p − 2)u³/3! + … for the modulus of u < 1. A nonnegative integer p makes the coefficients eventually zero and the expression finite. Other rational powers generally produce an infinite series; truncation is an approximation.
Negative exponent example — For p = −1, (1 + u)⁻¹ = 1 − u + u² − u³ + … for the modulus of u < 1. This is the geometric series with ratio −u. It differs from the finite polynomial identity obtained for a positive integer exponent, so do not apply it unrestricted at u = 2.
Square-root expansion — For p = 1/2, (1 + u)¹ᐟ² = 1 + u/2 − u²/8 + u³/16 + … for the modulus of u < 1. For p = −1/2, the first four terms are 1 − u/2 + 3u²/8 − 5u³/16. Recompute the falling coefficient products to check signs instead of assuming every fractional expansion alternates in the same way.
Factor out a constant — To expand √(4 + x), write 2(1 + x/4)¹ᐟ². The first four terms are 2 + x/4 − x²/64 + x³/512. The standard expansion requires the modulus of x/4 < 1, so −4 < x < 4; a function being real at some other input does not make this particular series valid there.
Numerical approximation — For √4.04 set x = 0.04 in that expansion, giving 2 + 0.01 − 0.000025 + 0.000000125 = 2.009975125 through the cubic term. The input satisfies the convergence condition. Use an approximation sign and compare with a calculator check; retaining three terms is not a universal guarantee of a specified number of decimal places.
Validity intersections — If an expression combines expansions in x and 2x, both the modulus of x < 1 and the modulus of 2x < 1 are required, so the common validity interval is the modulus of x < 1/2. State the intersection, not the wider union. Boundary convergence needs separate justification and is not implied by the strict standard binomial condition.
The two horizontal intervals use the same scale of 120 pixels per unit, with their common origin centred at x=0. Open endpoint circles exclude the boundaries. Requiring both |x|<1 and |2x|<1 selects their intersection −1/2<x<1/2, the green narrower interval. The rows are vertically separated only to compare the two conditions.
Partial-fraction series — Decompose (3x + 1)/((1 − x)(1 + 2x)) as (4/3)/(1 − x) − (1/3)/(1 + 2x). Expand the first denominator as 1 + x + x² + x³ + … and the second as 1 − 2x + 4x² − 8x³ + … . Combining gives 1 + 2x + 0x² + 4x³ + … with common validity the modulus of x < 1/2.
Truncation and checking — A truncated series retains only selected low-order terms; omitted terms can affect the requested precision. Check coefficients by symbolic expansion or recombination, track the expansion variable and assess whether its actual substituted magnitude is small enough. Near the validity boundary convergence can be slow, so validity alone does not mean a short truncation is accurate.

Discrete models timing assumptions and limitations

Additive and percentage models — A fictional scheme adding £20 to each monthly contribution is arithmetic in the contribution amount, whereas multiplying a balance by 1.04 each year gives geometric balance growth. A contribution sequence and a balance sequence are different objects. State exactly what uₙ measures and when additions or growth occur.
Initial-value convention — A fictional £500 balance with annual factor 1.04 is Bₙ = 500(1.04)ⁿ when n counts completed years and B₀ = 500. If the same observations are labelled u₁,u₂,… with u₁ initial, then uₙ = 500(1.04)ⁿ⁻¹. Ten completed years correspond to B₁₀ or u₁₁, not u₁₀.
Recurring contributions — For a fictional account with 1% growth applied each period before a £100 addition, Bₙ₊₁ = 1.01Bₙ + 100. If the addition happens before growth, the rule becomes 1.01(Bₙ + 100), which is different. Establish timing before deciding the first geometric term in any total.
Affine recurrence and fixed point — For u₁ = 2 and uₙ₊₁ = uₙ/2 + 3, a possible limiting fixed point L, a value left unchanged by the recurrence rule, satisfies L = L/2 + 3, giving 6. Subtracting 6 yields uₙ₊₁ − 6 = (uₙ − 6)/2, so uₙ = 6 − 4(1/2)ⁿ⁻¹. This proves convergence toward 6 and increasing terms here; solving only the fixed-point equation would not establish convergence in general.
For the stated initial value and recurrence, subtracting the fixed point yields a geometric deviation with ratio 1/2 and initial deviation −4. The explicit formula therefore approaches 6 from below. Solving L=f(L) alone would not prove convergence for a different recurrence or initial condition.
Finite physical time — A bouncing-height model with successive maximum heights hₙ = 2(0.6)ⁿ⁻¹ approaches zero mathematically. It ignores changing deformation, air resistance and the threshold at which a real bounce is no longer distinct. If modelling total distance, the first drop and later up-and-down legs need different treatment rather than summing heights blindly.
Threshold decisions — A model asking when a quantity first exceeds a threshold needs an integer time-step decision after solving an equation or inequality. For growth, use the smallest allowed integer meeting the actual condition; for decreasing values the inequality direction may change after logs. Substitute neighbouring indices to check the claimed first step.
Model evaluation — A recurrence reproducing early observations may fail as its assumed constant difference, percentage factor or external conditions change. Compare later observations, consider uncertainty and explain a specific limitation before proposing a refinement. The account values here are fictional mathematical examples, not current product rates or financial advice.
Interpreting infinite sums — An infinite mathematical total may provide an ideal limiting estimate even though a real process has finitely many meaningful steps. State convergence conditions and what is being summed, including any initial term, repeated leg or delayed contribution. Individual terms approaching zero is necessary for an ordinary infinite series to converge, but is not sufficient for arbitrary non-geometric series.

Test yourself

50 questions · Random sets of 10. These quick checks support revision; practise longer explanations and justified judgements too.