SOLUTION: Use the approach in Gauss's Problem to find the following sums of arithmetic sequences (do not use formulas): a. 1+2+3+4+...+49 b. 1+3+5+7+...+2009 c. 6+12+18+...+600 d. 1000

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Question 987074: Use the approach in Gauss's Problem to find the following sums of arithmetic sequences (do not use formulas):
a. 1+2+3+4+...+49
b. 1+3+5+7+...+2009
c. 6+12+18+...+600
d. 1000+995+990+...+5
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Gauss's method:  (average term) × (number of terms).

1. To find the average term, add the first and last term, and divide by 2.

2. To find the number of terms, subtract first and last terms, divide by
difference between terms and add 1.

a. 1+2+3+4+...+49

Average term: 1+49=50, divide by 2, get 25
Number of terms: 49-1=48, 2-1=1, 48÷1=48, 48+1=49
                 [Actually you could easily just tell there were 49 terms,
                  but that's not as easy on the others]
Average term × Number of terms = 25×49 = 1225  


b. 1+3+5+7+...+2009

Average term: 1+2009=2010, divide by 2, get 1005
Number of terms: 2009-1=2008, 3-1=2, 2008÷2=1004, 1004+1=1005 terms
Average term × Number of terms = 1005×1005 = 1010025 

c. 6+12+18+...+600

Average term: 6+600=606, divide by 2, get 303
Number of terms: 600-6=594, 12-6=6, 594÷6=99, 99+1=100 terms
Average term × Number of terms = 303×100 = 30300 



d. 1000+995+990+...+5

Average term: 1000+5=1005, divide by 2, get 502.5
Number of terms: 1000-5=995, 1000-995=5, 995÷5=199, 199+1=200 terms
Average term × Number of terms = 502.2×200 = 100500

Edwin
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