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Find all n satisfying 1 + 3 + 5 + 7 + ... + 125 = n + (n + 2) + (n + 4) + (n + 6) + ... + 209.

 Nov 15, 2019
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"Find all n satisfying 1 + 3 + 5 + 7 + ... + 125 = n + (n + 2) + (n + 4) + (n + 6) + ... + 209."

 

L = 1+3+5+7+...+125 = 3969

 

R = n+(n+2)+(n+4)+(n+6)+...+209

Write the last term of R as n+2N

R = n*(N+1) + N*(N+1)

 

Now the last term n + 2N = 209, so N = (209 - n)/2, hence:

R = (n + 209)(211 - n)/4

 

Equate L and R:

(n + 209)(211 - n)/4 = 3969

 

I'll leave you to solve this quadratic to find the two values of n.

 Nov 15, 2019

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