SOLUTION: I'm struggling with this one. Let f(n)=1^2+2^2+3^2+...+n^2. Thus, f(1)=1^2 f(2)=1^2+2^2 f(3)=1^2+2^2+3^2 f(4)=1^2+2^2+3^2+4^2 etc. It turns out that f is a polynomial o

Question 287961: I'm struggling with this one.
Let f(n)=1^2+2^2+3^2+...+n^2.
Thus,
f(1)=1^2
f(2)=1^2+2^2
f(3)=1^2+2^2+3^2
f(4)=1^2+2^2+3^2+4^2
etc.
It turns out that f is a polynomial of degree 3 in n. Figure out the coefficients of f:
f(n)=__n^3+__n^2+__n+__.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I'm struggling with this one.
Let f(n)=1^2+2^2+3^2+...+n^2.
Thus,
f(1)=1^2
f(2)=1^2+2^2
f(3)=1^2+2^2+3^2
f(4)=1^2+2^2+3^2+4^2
etc.
It turns out that f is a polynomial of degree 3 in n. Figure out the coefficients of f:
f(n)=__n^3+__n^2+__n+__.
------------------------------
Number pairs:
1::::1
2::::1+4 = 5
3::::1+4+9 = 14
4::::14+16 = 30
5::::30+25 = 55
6::::55+36 = 91
etc.
----
I ran a Cubic Regression program against
these pairs and got the following:
f(n) = (1/3)n^3 + (1/2)n^2 + (833/5000)n^2
==============================================
Cheers,
Stan H.
==============================================
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