11. Sequences and sum formulas
x counts TERMS, not values: sum of first x evens = x(x+1), first x odds = x^2.
Core ideas
- The counting formula counts TERMS, not values: count = (last - first)/step + 1. The +1 is where careless errors live.
- Sum of an evenly spaced run = (number of terms) * (first + last)/2.
- Shortcuts: the sum of the first x even integers is x(x+1); the sum of the first x odd integers is x^2. In both, x counts how many terms, not how big they get.
- For a recursive sequence, just crank out terms; patterns usually repeat within 4 to 6 steps.
Worked example 1
What is the sum of the first 25 positive even integers?
Show solution
The formula x(x+1) uses x = number of terms, so x = 25 (not 50, even though the terms run up to 50). Sum = 25 * 26 = 650. Sanity check with the general formula: 25 terms, first 2, last 50, sum = 25 * (2 + 50)/2 = 25 * 26 = 650. Same answer.
Worked example 2
How many multiples of 3 are there between 20 and 100, and what is their sum?
Show solution
First multiple of 3 past 20 is 21; last one before 100 is 99. Count = (99 - 21)/3 + 1 = 26 + 1 = 27 terms. Sum = count * average of endpoints = 27 * (21 + 99)/2 = 27 * 60 = 1620.
Practice set
Question 1
In an arithmetic sequence, the first term is 5 and each term after the first is 3 more than the previous term. What is the 20th term?
From the first term to the 20th term you take 19 steps, not 20.
The nth term is 5 + 3(n - 1). For n = 20: 5 + 3(19) = 5 + 57 = 62. Choice D, 65, comes from adding 3 twenty times instead of nineteen.
Question 2
What is the sum of the first 20 positive even integers?
There is a closed formula for the sum of the first x even integers.
The sum of the first x positive even integers is x(x + 1). With x = 20, the sum is 20 * 21 = 420. Choice A, 200, comes from averaging incorrectly; 400 comes from x^2, the odd-integer formula applied to evens.
Question 3
What is the sum of all even integers from 40 to 100, inclusive?
This range does not start at 2, so subtract one full-range sum from another, or count terms and use the average.
Evens 2 to 100 are the first 50 evens, summing to 50 * 51 = 2550. Evens 2 to 38 are the first 19 evens, summing to 19 * 20 = 380. The difference is 2550 - 380 = 2170. Alternatively, there are 31 terms with average (40 + 100)/2 = 70, and 31 * 70 = 2170. Choice A comes from using 30 terms, an off-by-one in the count.
Question 4
For a positive integer n, the sum of the first n positive even integers exceeds the sum of the first n positive odd integers by 35. What is n ?
Compare the two closed-form sums; the difference simplifies to something very small.
Sum of first n evens is n(n + 1) = n^2 + n. Sum of first n odds is n^2. The difference is n^2 + n - n^2 = n. Each even integer is exactly 1 more than the odd integer paired with it, so the gap after n terms is n. Since the difference is 35, n = 35.