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Question 18128: f(n)=1^2+2^2+3^2+...+n^2 it turns out that f is a polynomial of degree x in n. Figure out the coefficients of f: f(n)=___n^3+___n^2+___n+___.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! find...f(n)=1^2+2^2+3^2+...+n^2
consider the identity...
n(n+1)(2n+1)-(n-1)n(2n-1)=6n^2..(you can expand the 2 expressions,simplify to get the result..n(n+1)(2n+1)-(n-1)n(2n-1)=(n^2+n)(2n+1)-(n^2-n)(2n-1)
=(2n^3+n^2+2n^2+n)-(2n^3-n^2-2n^2+n)=2n^3+3n^2+n-2n^3+3n^2-n=6n^2)
now put n=1,2,3...etc in the above and add up as shown below
..................n(n+1)(2n+1)-(n-1)n(2n-1)=6n^2
for n=1...........1.2.3-0..................... =6*1^2
for n=2...........2.3.5-1.2.3..................=6*2^2
for n=3...........3.4.7-2.3.5..................=6*3^2
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for n=n-1.........(n-1)n(2n-1)-(n-2)(n-1)(2n-3)=6*(n-1)^2
for n=n............n(n+1)(2n+1)-(n-1)n(2n-1)...=6n^2
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adding all above we get......... n(n+1)(2n+1)...=6*(1^2+2^2+3^2+...+n^2)
hence we have f(n)=(1^2+2^2+3^2+...+n^2)=n(n+1)(2n+1)/6
or..f(n)=(1/3)n^3+(1/2)n^2+(1/6)n
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