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23 Nov 2022, 11:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:16) correct
27%
(02:23)
wrong
based on 123
sessions
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Date
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Not Attempted Yet
\(1+2+3+......+n=\frac{n(n+1)}{2}\)
For each integer n greater than 1, the sum of the first n positive integers is given by the formula shown.
If the average (arithmetic mean) of the first n positive integers is k, what is the sum of the first n positive integers in terms of k?
A. \(k^2−k\)
B. \(k^2−k^2\)
C. \(k^2+k^2\)
D. \(2k^2−k\)
E. \(4k^2+2k\)
For each integer n greater than 1, the sum of the first n positive integers is given by the formula shown.
If the average (arithmetic mean) of the first n positive integers is k, what is the sum of the first n positive integers in terms of k?
A. \(k^2−k\)
B. \(k^2−k^2\)
C. \(k^2+k^2\)
D. \(2k^2−k\)
E. \(4k^2+2k\)
23 Nov 2022, 11:28
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Average= (n+1)/2 =k
n=2k-1
Now substitute n=2k-1 in n(n+1)/2.
(2k-1)(2k-1+1)/2
(2k-1)k
2k^2-k
Hence D is the correct choice.
Posted from my mobile device
n=2k-1
Now substitute n=2k-1 in n(n+1)/2.
(2k-1)(2k-1+1)/2
(2k-1)k
2k^2-k
Hence D is the correct choice.
Posted from my mobile device
23 Nov 2022, 11:32
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{n(n+1)}/2n=k
(n+1)/2=k
n=2k-1
Sum of the first n positive integers in terms of k:
(2k-1)(2k-1+1)/2= (2k-1)(2k)/2
= 4k^2-2k/2= 2k^2-k
Answer is option D
Posted from my mobile device
(n+1)/2=k
n=2k-1
Sum of the first n positive integers in terms of k:
(2k-1)(2k-1+1)/2= (2k-1)(2k)/2
= 4k^2-2k/2= 2k^2-k
Answer is option D
Posted from my mobile device
