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How do you find sum of the first 100 odd integers? - Answers

Suppose t(n) is the nth term. Then t(n) = -1 + 2n (for n = 1, 2, 3, ... 100) Now t(1) + t(100) = (-1 + 2*1) + (-1 +2*100) = 1 + 199 = 200 t(2) + t(99) = (-1 + 2*2) + (-1 +2*99) = 3 + 197 = 200 there are 50 such pairs. So the grand total is 50*200 = 10000



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How do you find sum of the first 100 odd integers? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_find_sum_of_the_first_100_odd_integers

Suppose t(n) is the nth term. Then t(n) = -1 + 2n (for n = 1, 2, 3, ... 100) Now t(1) + t(100) = (-1 + 2*1) + (-1 +2*100) = 1 + 199 = 200 t(2) + t(99) = (-1 + 2*2) + (-1 +2*99) = 3 + 197 = 200 there are 50 such pairs. So the grand total is 50*200 = 10000



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https://math.answers.com/math-and-arithmetic/How_do_you_find_sum_of_the_first_100_odd_integers

How do you find sum of the first 100 odd integers? - Answers

Suppose t(n) is the nth term. Then t(n) = -1 + 2n (for n = 1, 2, 3, ... 100) Now t(1) + t(100) = (-1 + 2*1) + (-1 +2*100) = 1 + 199 = 200 t(2) + t(99) = (-1 + 2*2) + (-1 +2*99) = 3 + 197 = 200 there are 50 such pairs. So the grand total is 50*200 = 10000

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      Suppose t(n) is the nth term. Then t(n) = -1 + 2n (for n = 1, 2, 3, ... 100) Now t(1) + t(100) = (-1 + 2*1) + (-1 +2*100) = 1 + 199 = 200 t(2) + t(99) = (-1 + 2*2) + (-1 +2*99) = 3 + 197 = 200 there are 50 such pairs. So the grand total is 50*200 = 10000

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