Sum of Sequential Integers (featuring Swiss Micros DM32)
What is the sum of the following:
100 + 101 + 102 + 103 + 104 + 105 + … + 997 + 998 + 999?
It’s doable on a calculator, but going straight forward will require a lot of keystrokes (unless you have access to the sigma function (Σ)).
Note that:
100 + 101 + 102 + 103 + 104 + 105 + … + 997 + 998 + 999
= 100 + (100 + 1) + (100 + 2) + (100 + 3) + (100 + 4) + …. + (100 + 897) + (100 + 898) + (100 + 899)
= (100 + 0) + (100 + 1) + (100 + 2) + (100 + 3) + (100 + 4) + …. + (100 + 897) + (100 + 898) + (100 + 899)
Notice the sum starts with a base, 100. The sequence goes for 900 terms in the form of 100+n where n = 0 to 899. Let’s look at a general case.
Let x be a base integer, and S be the sum as:
S = (x + 0) + (x + 1) + (x + 2) + (x + 3) + …. + (x + n)
Rearranging the terms leads us to:
S = (x + x + x + x + … + x) + (0 + 1 + 2 + 3 + 4 + … + n)
There are n + 1 pairs:
S = (x * (n + 1)) + (0 + 1 + 2 + 3 + 4 + … + n)
S = (x * (n + 1)) + (1 + 2 + 3 + 4 + … + n)
The sum of Σ( k, k = 1 to k = n) = 1 + 2 + 3 + 4 + … + n = n * (n + 1) / 2
Then:
S = (x * (n + 1)) + n * (n + 1) / 2
Let’s look at some specific examples:
n = 1
S = (x + 0) + (x + 1)
S = (x + x) + (0 +1)
S = 2 * x + 1
n = 2
S = (x + 0) + (x + 1) + (x + 2)
S = (x + x + x) + (0 + 1 + 2)
S = 3 * x + 3
2 * 3 / 2 = 3
n = 3
S = (x + 0) + (x + 1) + (x + 2) + (x + 3)
S = (x + x + x + x) + (0 + 1 + 2 + 3)
S = 4 * x + 6
3 * 4 / 2 = 6
S = 500 + 501 +502 + 503
S = (500 + 0) + (500 + 1) + (500 + 2) + (500 + 3)
S = (500 + 500 + 500 + 500) + (0 + 1 + 2 + 3)
S = ((3 + 1) * 500) + (3 * 4 / 2)
S = 2000 + 6
S = 2006
S = 250 + 251 + 252 + 253 + … + 269 + 270
Base = 250
n = 270 – 250 = 20
Then:
S = 250 * (20 + 1) + 20 * 21 / 2
S = 5460
Let’s go our original problem:
S = 100 + 101 + 102 + 103 + 104 + 105 + … + 997 + 998 + 999
Base = 100
n = 999 -100 = 899
Then:
S = 100 * (899 + 1) + 899 * 900 / 2
S = 90000 + 404550
S = 494550
The program code calculates the sum:
Swiss Micros DM32 Program (also HP 32SII)
S01 LBL S
S02 INPUT X
S03 INPUT N
S04 RCL N
S05 1
S06 +
S07 RCL× N
S08 RCL N
S09 1
S10 RCL+ N
S11 ×
S12 2
S13 ÷
S14 +
S15 RTN
Eddie
All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
