sum_{n=1}^5(2n-1) =1+3+5+7+9 =25

Notice that the expression (2n − 1) generates odd numbers. as n takes values 1, 2, 3, 4, 5.

If we want to generate even numbers, we would use 2n, as follows:

sum_{n=1}^5{2n} =2+4+6+8+10 =30

To generate alternate positive and negative numbers, we multiply the expression in the summation by (−1)n+1. For example:

sum_{n=1}^5(-1)^{n+1}(n) =1-2+3-4+5 =3

And here's a final example, giving us alternate positive and negative fractional terms with even denominators:

sum_{n=1}^5\frac{(-1)^{n+1}}{2n} =\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10} =\frac{47}{120}

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