Can we directly add up these percentages?

Algebra Level 3

∑ n = 0 100 ( − 1 ) n + 1 n % = x % \large \displaystyle \sum_{n=0}^{100} (-1)^{n+1} n \% = x\%

Find x x .


The answer is -50.

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Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Akshat Sharda
Dec 19, 2015

∑ n = 0 100 ( − 1 ) n + 1 n % ∑ n = 0 100 ( − 1 ) n + 1 n 100 1 100 ∑ n = 0 100 ( − 1 ) n + 1 n 1 100 ( 0 + 1 − 2 + 3 − 4 + 5 − … + 99 − 100 ) 1 100 ( − 50 ) = − 50 % \displaystyle \sum_{n=0}^{100} (-1)^{n+1} n \% \\ \displaystyle \sum_{n=0}^{100} (-1)^{n+1} \frac{n}{100} \\ \frac{1}{100} \displaystyle \sum_{n=0}^{100} (-1)^{n+1} n \\ \frac{1}{100}\left(0+1-2+3-4+5-\ldots+99-100\right) \\ \frac{1}{100}\left(-50\right)=-50 \%

9 Helpful 0 Interesting 0 Brilliant 0 Confused

That was a very interesting problem. I liked the straightforward approach.

Drex Beckman - 5 years, 5 months ago

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