Camsie is starting a new job where she earns *n* dollars on her* n*th day of work. What is the total amount that Camsie will earn for the first ten days of work at her new job?

*On each of the first 10 days of work at her new job, Camsie will earn 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 dollars, respectively. That’s a total of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 10(1 + 10)/2 = 5(11) = **55 **dollars.*

How many days will Camsie have worked at her new job when her cumulative earnings total 595 dollars?

*Camsie's cumulative earnings for working the first n days at her new job equal the sum of the integers from 1 to n. Since 1 + 2 + 3 + … + n = (n/2)(n + 1), we can solve the equation (n/2)(n + 1) = 595 *→

*n(n + 1) = 1190 →*

*n*→

^{2}+ n – 1190 = 0. Factoring the quadratic yields (n – 34)(n + 35) = 0. So, n – 34 = 0*n = 34 and n + 35 = 0*→

*n = –35. Therefore, Camsie’s cumulative earnings will total 595 dollars when she has worked*

*34**days.*

What is the minimum number of days Camsie must work at her new job for her cumulative earnings to exceed 1000 dollars?

*Using the same expression for the sum of the integers from 1 to n, we have the following inequality: (n/2)(n + 1) > 1000 *→ *n(n + 1) > 2000. Simplifying yields the quadratic expression n ^{2} + n – 2000, which cannot be factored. However, we are essentially*

*looking for two consecutive integers, n and n + 1, that have a product greater than 2000. Since 40*

^{2}= 1600 and 50^{2}= 2500, we know the integers are between 40 and 50. The square of 45 is 2025, so we check 44 × 45 and 45 × 46 and get 1980 and 2070, respectively. Therefore, we conclude that in order for Camsie’s cumulative earnings to exceed 1000 dollars, she must work at least

*45**days.*

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