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Find n so that 20 + 21 + 22 + ... + n = 2021.

 Dec 24, 2020
 #1
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+1

\(\sum_{20}^{66}\) n   = 2021

 Dec 24, 2020
 #2
avatar+114325 
+1

We can solve for n  thusly

 

Sum =   (first term + last term)  (number of terms) /2

 

Note.....number of terms  = n - 20 + 1  =  n -19

 

So  we  have that

 

(20 + n)  ( n - 19)/2  =  2021    multiply through by 2    and simplify

 

n^2 + n - 380   =  4042          subtract  4042 from both sides

 

n^2  + n - 4422   =   0  

 

We can factor this as

 

(n + 67)  ( n - 66)    =  0

 

The second factor set  =   0   gives us the positive answer we  need  for n

 

n - 66    =  0

 

n  =  66

 

 

cool cool cool

 Dec 24, 2020

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