If \(n\) is a positive integer, the sum of the integers from 1 to \(n,\) inclusive, equals \(\dfrac{n(n+1)}{2}.\) Which

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If \(n\) is a positive integer, the sum of the integers from 1 to \(n,\) inclusive, equals \(\dfrac{n(n+1)}{2}.\) Which of the following equals the sum of the integers from 1 to \(2n,\) inclusive?


A. \(n(n+1)\)

B. \(\dfrac{n(2n+1)}{2}\)

C. \(n(2n+1)\)

D. \(2n(n+1)\)

E. \(2n(2n+1)\)

[spoiler]OA=C[/spoiler]

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VJesus12 wrote:
Mon May 11, 2020 6:21 am
If \(n\) is a positive integer, the sum of the integers from 1 to \(n,\) inclusive, equals \(\dfrac{n(n+1)}{2}.\) Which of the following equals the sum of the integers from 1 to \(2n,\) inclusive?


A. \(n(n+1)\)

B. \(\dfrac{n(2n+1)}{2}\)

C. \(n(2n+1)\)

D. \(2n(n+1)\)

E. \(2n(2n+1)\)

[spoiler]OA=C[/spoiler]

Solution:

Replacing n by 2n in the formula, the sum of the integers from 1 to 2n, inclusive, is:

2n(2n + 1)/2 = n(2n + 1)

Answer: C

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