Download Summation Formulas: Arithmetic and Geometric Progressions and more Papers Mathematics in PDF only on Docsity! Math 101 Summation Formulas 1. Arithmetic Sums The computation S = 1 + 2 + 3 + 4 + 5 + · · · + 999 + 1000 could be done on a hand calculator by entering each number, starting with 1 and ending with 1000. The chances for making an error, by pressing a wrong key while entering a digit, is very great. Besides, it takes a long time to enter a thousand numbers! Here’s a better way: Write the list once, and directly below it, write the same list with the numbers in reverse order: S =1 + 2 + 3 + 4 + 5 + · · ·+ 999 + 1000 S =1000 + 999 + · · ·+ 5 + 4 + 3 + 2 + 1. Now add each entry in the top row with the entry in the row directly below it: 1+1000 = 1001, 2+999 = 1001, 3+998 = 1001, etc. You get 1000 sums or 1001. So 2S = 1000×1001 = 1001000. Dividing by two gives us the value S = 1001000/2 = 500500. Does this trick work for other sums? Exercise. Try the “backwards” method on: 1. 1 + 2 + 3 + 4 + 5 + · · ·+ 499 + 500 2. 1 + 3 + 5 + 7 + 9 + · · ·+ 95 + 97 + 99. 3. 11 + 12 + 13 + · · ·+ 98 + 99 + 100. 4. 5 + 10 + 15 + 20 + · · ·+ 190 + 195 + 200. 5. 2 + 5 + 8 + 11 + 14 + 17 + · · ·+ 992 + 995 + 998. Sequences like these are called arithmetic progressions. Note that we obtain each term by adding the same number to the previous term. 1 2 Exercise. 6. Find the following sums by taking the terms two at a time: 1− 3 + 5− 7 + 9− 11 + · · ·+ 93− 95 + 97− 99 7. Find the following sum of infinitely many numbers: 3 10 + 3 100 + 3 1000 · · · . Could you use a calculator to do this problem? 2. Geometric Sums A second type of sum we will meet in this class is obtained by multiplying the previous term by a common factor. These are called geometric progressions. If the first term is 1, then the sum of a geometric progression is S = 1+x+x2 +x3 + · · ·+xn−1 +xn. The trick for adding a geometric progression is to multiply S by x on a line just above S: xS =x+ x2 + x3 + · · ·+ xn−1 + xn + xn+1 S =1 + x+ x2 + x3 + · · ·+ xn−1 + xn. Now subtact the second line from the first and notice that almost every term in the first row cancels a term in the seond row. The only uncancelled terms are the xn+1 in line 1 and the 1 in line 2. Hence xS − S = xn+1 − 1 Since xS − S = (x− 1)S we have (x− 1)S = xn+1 − 1 Dividing by x− 1 gives us the value of S: S = xn+1 − 1 x− 1 Example. Compute the sum 1 + 2 + 4 + 8 + 16 + 32. Written as powers of 2 this sum is 1 + 2 + 4 + 8 + 16 + 32 = 1 + 21 + 22 + 23 + 24 + 25
