Exercise 66 Page 100

Consider a list of consecutive integers. 1, 2, 3, 4, 5, 6 No matter which integer we choose, the next integer is 1 greater than the one we chose. For this exercise, we want to find 3 consecutive integers whose sum is 45. first+second+third=45 If we call the first integer n, the two consecutive integers after it will be n+1 and n+2. n^(1^(st))+(n+1)^(2^(nd))+(n+2)^(3^(rd))=45 To find the first of the three integers, we will solve this equation for n. After we have the value of n, we will also be able to find the value of n+1 and n+2.

The first integer is 14. Now we can find what 3 consecutive integers add up to be 45.