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Find the sum \(1^2 + 2^2 + 3^2 + \dots + 1000^2.\)

 Feb 10, 2020
 #1
avatar+109064 
+1

Maybe easier ways to do this......but.....using the sum of differences we have 

 

1       5        14          30         55

    4       9          16          25

         5         7            9

               2         2

 

We have 3 non-zero results  so  we can express the progessive sums  as a cubic polynomial

an^3 + bn^2 + cn + d        where   n  is the sum of the first n squares

 

We have that

 

a(1)^3   + b(1)^2  + c(1) + d  =  1

a(2)^3  +  b(2)^2  + c(2) + d = 5

a(3)^3  + b(3)^2 + c(3) + d  =  14

a(4)^3  + b(4)^2  + c(4) + d =  30

 

 

a  + b  +  c  + d  =  1

8a + 4b + 2c+ d  = 5

27a + 9b + 3c + d = 14

64a + 16b + 4c + d = 30

 

The  solution to this  is   

 

a = 1/3    b  = 1/2     c  =1/6    d   = 0

 

So.....the resulting polynomial is

 

(1/3)n^3  + (1/2)n^2 + (1/6)n

 

So.....the sum  of the first 1000 squares is

 

(1/3) (1000)^3  + (1/2)(1000)^2  + (1/6)(1000)   = 

 

333,833,500

 

 

cool cool cool

 Feb 10, 2020
edited by CPhill  Feb 10, 2020
 #2
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0

CPhill: Did you forget the formula for the "sum of consecutive squares"??

 

n = 1000; a = 1/6 * (2*n + 1)*n*(n + 1) =333,833,500

 Feb 10, 2020
 #3
avatar+109064 
0

Yeah....I knew there was a formula but I forgot exactly what it was......!!!

 

 

cool cool cool

CPhill  Feb 10, 2020

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