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Find n if (1^2 + 2^2 + ... + n^2)/(1 + 2 + .. + n) = 43/3.

 Dec 14, 2019
 #1
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Solve for n:
(2 n + 1)/3 = 43/3

Multiply both sides of (2 n + 1)/3 = 43/3 by 3:
1 (2 n + 1) = 3×43/3

3×43/3 = (3×43)/3:
(3 (2 n + 1))/3 = (3×43)/3

(3 (2 n + 1))/3 = 3/3×(2 n + 1) = 2 n + 1:
2 n + 1 = (3×43)/3

(3×43)/3 = 3/3×43 = 43:
2 n + 1 = 43


n = 21

 Dec 14, 2019
 #2
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To clarify the above answer, consider this:

 

Sum of squares of consecutive numbers =1/6[2n +1) * n * (n +1)..........(1)

 

Sum of consecutive numbers                     =1/2[n * (n +1)].......................(2)

 

Divide (1) by (2) and you get:                     = 1 /3 *(2n +1)

 

1/3[2n +1] =43/3           Cross multiply

 

2n +1 = 43

 

Therefore: n = 21

 Dec 14, 2019

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