The ratio of the sums of the first n even num | vedclass

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Question

(C) The sum of the first $n$ even numbers is given by the arithmetic progression $2, 4, 6, \dots, 2n$.Sum $S_{E} = \frac{n}{2}(2 + 2n) = n(n + 1)$.The

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$(n + 1):n$

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(C) The sum of the first $n$ even numbers is given by the arithmetic progression $2, 4, 6, \dots, 2n$.Sum $S_{E} = \frac{n}{2}(2 + 2n) = n(n + 1)$.The sum of the first $n$ odd numbers is given by the arithmetic progression $1, 3, 5, \dots, (2n - 1)$.Sum $S_{O} = \frac{n}{2}(1 + 2n - 1) = \frac{n}{2}(2n) = n^2$.The ratio of the sums is $\frac{S_{E}}{S_{O}} = \frac{n(n + 1)}{n^2} = \frac{n + 1}{n}$.Thus,the ratio is $(n + 1):n$.

The ratio of the sums of the first $n$ even numbers and $n$ odd numbers is:

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