Addition Techniques

Let’s say you want to add a set of consecutive natural numbers from 1 to n.  Instead of taking the time to add every number together the sum can be found be using the formula below.

Consider the case where n is even.  Now take the nth number and 1.  Their sum is n+1.  Now shift the numbers by adding one to the lower number and subtracting one from the higher number.  So now (1+1)+(n-1)= n+1 + (1-1) = n+1.  In general, when n is even there are n/2 pairs of numbers which sum up to n+1.  Therefore the sum of natural numbers from 1 to n is 1/2*n*(n+1).

When n is odd take the (n-1)st number and one.  Their sum is n.  Now add one to the lower number and subtract one from the higher number.  Then (1+1) +(n–1) -1 = n + (2-2) = n.  By taking numbers in the sum in pairs there are (n-1)/2 pairs of numbers which sum to n; plus n itself.  Therefore the sum can be written as the sum of (n+1)/2 numbers with a value of n.  Thus, the sum from 1 to n is 1/2*n*(n+1).

Examples:

Sum the numbers from one to ten.  The pairs which sum to eleven are: (1,10), (2,9), (3,8), (4,7), and (5,6).  There are 10/2 = 5 pairs of these numbers.  Therefore, the sum is 1/2*10*11 = 55.

Sum the numbers from one to eleven.  The pairs which sum to eleven are: (1,10), (2,9), (3,8), (4,7), and (5,6).  There are 10/2 = 5 pairs of these numbers, and then there is eleven itself.  Then the sum can be written as 11+11+11+11+11+11 or the sum of (n+1)/2=6 numbers whose value is n.  Therefore, the sum is 1/2*11*12 = 66.