Sequences, Series, and the Binomial Theorem

Sequences and series turn up all over the Mathematics, Surveying and Transportation Engineering (MSTE) paper: a settling foundation whose annual movement shrinks by a fixed ratio, a payment plan that grows by a fixed amount, a repeating decimal you must turn into an exact fraction, or a quantity you must estimate with a few terms of a series when no calculator function will do. This lesson drills the patterns the board expects you to recognize on sight, arithmetic, geometric, and harmonic progressions, then the binomial theorem and a first look at power series, with every computation carried to a finished number.

Arithmetic sequences and series

An arithmetic sequence (AP) advances by adding the same common difference d at every step. If the first term is a1, then the nth term and the sum of the first n terms are:

a_n = a1 + (n - 1)d

S_n = (n/2)(a1 + a_n) = (n/2)[2a1 + (n - 1)d]

The two sum formulas are the same thing: use the first when you already know the last term a_n, and the second when you only know a1, d, and n. The single most common slip is writing n instead of (n - 1) in the nth-term formula, which shifts every answer by one d.

Worked example: An AP has first term a1 = 3 and common difference d = 4. The 10th term is a10 = 3 + (10 - 1)(4) = 3 + 36 = 39. The sum of the first 10 terms is S_10 = (10/2)(3 + 39) = 5(42) = 210. If instead you know only a1, d, and n, the packed formula gives the same value: S_10 = (10/2)[2(3) + 9(4)] = 5[6 + 36] = 5(42) = 210.

Geometric sequences and series

A geometric sequence (GP) advances by multiplying by the same common ratio r at every step. Its nth term and the sum of the first n terms are:

a_n = a1 x r^(n - 1)

S_n = a1(r^n - 1)/(r - 1), valid when r is not 1

The exponent is (n - 1), not n, for the same reason as the AP: the first term has been multiplied by r zero times. When r is between 0 and 1 the terms shrink; when r is greater than 1 they grow; when r is negative the terms alternate in sign.

Worked example: A GP has a1 = 5 and r = 2. The 6th term is a6 = 5 x 2^(6 - 1) = 5 x 32 = 160. For a1 = 3 and r = 2, the sum of the first 5 terms is S_5 = 3(2^5 - 1)/(2 - 1) = 3(32 - 1)/1 = 3(31) = 93, which you can confirm by adding 3 + 6 + 12 + 24 + 48 = 93.

The sum to infinity of a geometric series

When the common ratio satisfies |r| < 1, each term is smaller than the last and the partial sums settle toward a finite limit. That limit is the sum to infinity:

S_infinity = a1 / (1 - r), valid only when |r| < 1

If |r| >= 1 the terms do not shrink and the series diverges, so no finite sum exists. This is the formula behind converting a repeating decimal into an exact fraction.

Worked example (series): For 12 + 6 + 3 + 1.5 + ... the first term is a1 = 12 and the ratio is r = 1/2, so S_infinity = 12 / (1 - 0.5) = 12 / 0.5 = 24. Worked example (repeating decimal): Read 0.4444... as the geometric series 0.4 + 0.04 + 0.004 + ..., with a1 = 0.4 and r = 0.1. Then S = 0.4 / (1 - 0.1) = 0.4 / 0.9 = 4/9, the exact fraction.

Harmonic sequences and the three means

A harmonic sequence (HP) is one whose reciprocals form an arithmetic sequence. So to find any term of an HP, flip every term to build the AP of reciprocals, work on that AP with the ordinary formulas, then flip the answer back. There is no simple closed form for the sum of a harmonic series, so board items ask instead for a particular term or for the harmonic mean.

The harmonic mean is the third of the classic means. For two positive numbers a and b:

• Arithmetic mean: AM = (a + b) / 2
• Geometric mean: GM = sqrt(ab)
• Harmonic mean: HM = 2ab / (a + b)

These always obey AM >= GM >= HM, with equality only when a = b, and they are tied together by GM^2 = AM x HM, a fast way to check your work.

Mean	Formula for a and b	Value for 3 and 6
Arithmetic (AM)	(a + b) / 2	4.5
Geometric (GM)	sqrt(ab)	4.243
Harmonic (HM)	2ab / (a + b)	4

Worked example (HP term): Find the 6th term of 1/2, 1/5, 1/8, 1/11, ... The reciprocals 2, 5, 8, 11 form an AP with a1 = 2 and d = 3, so the 6th reciprocal is 2 + (6 - 1)(3) = 17, and the 6th term is 1/17. Worked example (means): For 3 and 6, HM = 2(3)(6) / (3 + 6) = 36 / 9 = 4, while AM = 4.5 and GM = sqrt(18) = 4.243. The check holds: GM^2 = 18 and AM x HM = 4.5 x 4 = 18.

The binomial theorem and the general term

The binomial theorem expands any power of a two-term sum:

(a + b)^n = C(n,0) a^n + C(n,1) a^(n-1) b + C(n,2) a^(n-2) b^2 + ... + C(n,n) b^n

where the binomial coefficient is C(n,k) = n! / [k!(n - k)!]. Three facts fall straight out of this: the expansion has n + 1 terms, the sum of all the coefficients is found by setting a = b = 1 so it equals (1 + 1)^n = 2^n, and each row of Pascal's triangle simply lists the coefficients for that power.

Often you want only one term, not the whole expansion. The general term, counting r from 0, is:

T_(r+1) = C(n, r) a^(n - r) b^r

To hit a required power of the variable, solve for the r that produces that power, then evaluate that single term. This saves you from writing out an eight-term expansion just to read off one coefficient.

Worked example (full expansion): (x + 2)^4 = x^4 + C(4,1) x^3 (2) + C(4,2) x^2 (2)^2 + C(4,3) x (2)^3 + (2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16. Worked example (one term): Find the term containing x^2 in (2x + 3)^5. Here a = 2x and b = 3, and the power of x is (5 - r), so set 5 - r = 2 to get r = 3. Then T_4 = C(5,3)(2x)^(5-3)(3)^3 = 10 x (2x)^2 x 27 = 10 x 4x^2 x 27 = 1080 x^2, so the coefficient is 1080.

A first look at power series

A power series adds up infinitely many powers of x with fixed coefficients, and the geometric series is its prototype:

1 / (1 - x) = 1 + x + x^2 + x^3 + ..., valid for |x| < 1

The set of x for which a power series converges is an interval around 0 (its radius of convergence); outside that interval the terms do not shrink and the sum is meaningless. Several standard functions have series you can use to approximate a value by hand: keep the first few terms and the error is small when x is small.

Function	Power series	Converges for
1 / (1 - x)	1 + x + x^2 + x^3 + ...	|x| < 1
e^x	1 + x + x^2/2! + x^3/3! + ...	all real x
(1 + x)^p (binomial series)	1 + p x + [p(p - 1)/2!] x^2 + ...	|x| < 1

Worked example (exponential): Estimate e^0.2 from the first three terms of e^x: 1 + 0.2 + (0.2)^2/2 = 1 + 0.2 + 0.04/2 = 1 + 0.2 + 0.02 = 1.22, close to the true 1.2214. Worked example (binomial series): Estimate sqrt(1.02) with (1 + x)^(1/2) approximately 1 + x/2 for small x, taking x = 0.02: 1 + 0.02/2 = 1 + 0.01 = 1.01, close to the true 1.00995.

Exam-day strategy

• Classify the sequence first: constant difference between terms means arithmetic, constant ratio means geometric, and reciprocals that form an AP means harmonic. The formula you reach for follows from that one decision.
• Watch the exponent and the count: the nth term uses (n - 1), not n. Writing the wrong one shifts every AP and GP answer by one step, the most common error on these items.
• Only use S_infinity = a1 / (1 - r) after you confirm |r| < 1; if |r| >= 1 the series diverges and there is no finite sum.
• Turn a repeating decimal into a geometric series (a1 is the repeating block as a decimal, r is a power of 1/10) and apply the sum-to-infinity formula to get the exact fraction.
• For one coefficient in a binomial expansion, use the general term T_(r+1) = C(n, r) a^(n-r) b^r: solve for the r that gives the required power instead of expanding the whole thing.
• Check binomial work against Pascal's triangle and remember the shortcuts: n + 1 terms, and the coefficients sum to 2^n.
• For a quick hand estimate, keep two or three terms of e^x or the binomial series (1 + x)^p; the smaller x is, the better the approximation.