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How do you add numbers 1 tru 100? - Answers

There are two ways to solve this problem: You may notice that if you take away 100 from the series and add the smallest and largest number (1 + 99), you get 100. Then if add the next numbers in (2 + 98), you also get 100. Soon, you see a pattern. 3 + 97 = 100, 4 + 96 = 100 and so on. If you do this all the way to 50, you get 100, forty-nine time or 49*100 which 4900. You then add the 100 you took away earlier and add that to 4900 to get 5000. Lastly, because there is only one 50 in 1 to 100, it stays as 50 and you add that to 5000 to get the answer: 5050. The second way is to use the summation equation. This only works if you are starting from 1 and adding all the consecutive numbers to a certain number, say n. The equation is n(n+1)/2 where "n" is the final number in the series which in this case is 100. So 100(100+1)/2 = 100(101)/2 = 10100/2 = 5050.



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How do you add numbers 1 tru 100? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_add_numbers_1_tru_100

There are two ways to solve this problem: You may notice that if you take away 100 from the series and add the smallest and largest number (1 + 99), you get 100. Then if add the next numbers in (2 + 98), you also get 100. Soon, you see a pattern. 3 + 97 = 100, 4 + 96 = 100 and so on. If you do this all the way to 50, you get 100, forty-nine time or 49*100 which 4900. You then add the 100 you took away earlier and add that to 4900 to get 5000. Lastly, because there is only one 50 in 1 to 100, it stays as 50 and you add that to 5000 to get the answer: 5050. The second way is to use the summation equation. This only works if you are starting from 1 and adding all the consecutive numbers to a certain number, say n. The equation is n(n+1)/2 where "n" is the final number in the series which in this case is 100. So 100(100+1)/2 = 100(101)/2 = 10100/2 = 5050.



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https://math.answers.com/math-and-arithmetic/How_do_you_add_numbers_1_tru_100

How do you add numbers 1 tru 100? - Answers

There are two ways to solve this problem: You may notice that if you take away 100 from the series and add the smallest and largest number (1 + 99), you get 100. Then if add the next numbers in (2 + 98), you also get 100. Soon, you see a pattern. 3 + 97 = 100, 4 + 96 = 100 and so on. If you do this all the way to 50, you get 100, forty-nine time or 49*100 which 4900. You then add the 100 you took away earlier and add that to 4900 to get 5000. Lastly, because there is only one 50 in 1 to 100, it stays as 50 and you add that to 5000 to get the answer: 5050. The second way is to use the summation equation. This only works if you are starting from 1 and adding all the consecutive numbers to a certain number, say n. The equation is n(n+1)/2 where "n" is the final number in the series which in this case is 100. So 100(100+1)/2 = 100(101)/2 = 10100/2 = 5050.

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      There are two ways to solve this problem: You may notice that if you take away 100 from the series and add the smallest and largest number (1 + 99), you get 100. Then if add the next numbers in (2 + 98), you also get 100. Soon, you see a pattern. 3 + 97 = 100, 4 + 96 = 100 and so on. If you do this all the way to 50, you get 100, forty-nine time or 49*100 which 4900. You then add the 100 you took away earlier and add that to 4900 to get 5000. Lastly, because there is only one 50 in 1 to 100, it stays as 50 and you add that to 5000 to get the answer: 5050. The second way is to use the summation equation. This only works if you are starting from 1 and adding all the consecutive numbers to a certain number, say n. The equation is n(n+1)/2 where "n" is the final number in the series which in this case is 100. So 100(100+1)/2 = 100(101)/2 = 10100/2 = 5050.
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