sum of first N integers is N*(N+1)/2 but what is sum of first N squares ie 1+4+9+...+N^N?

Answer 1

Sum on n² is n(n+1)(2n+1)/6
for n^3 is (n(n+1)/2)²
for n^4 is n(n+1)(2n+1)(3n²+3n-1)/30

There's a very elegant solution for sum on n^m (where m is any integer) that involves the Bernoulli numbers, but I don't remember it off the top of my head and I'm *not* gonna get up and go look for it until I have a couple more cups of coffee and another cigarette or two in me

You can probably find it on the Web.

And.... Did you know that the sum of the first n odd numbers is n² ? e.g. 1+3+5+7+9 = 25 = 5² or
1+3+5+7+9+11+13+15+17+19 = 100 = 10²
Is that too Kewl, or what ☺ (See if you can figure out why it works that way. Hint: use induction)


Doug

Answer 2

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RE:
sum of first N integers is N*(N+1)/2 but what is sum of first N squares ie 1+4+9+...+N^N?

Answer 3

Let's express it as a function .
f(n)-f(n-1)=n*n => f(n) takes n to the power of three since n*n is
f(0) = 0
n to th power of 2 .
f(n)=a*n*n*n+b*n*n+c*n
f(n-1) =a*n*n*n -a+3a*n-3a*n*n+b*n*n+b-2*n*b+c*n-c
f(n)-f(n-1)=n*n
3a*n-3a*n*n-2*n*b-c-a+b=-n*n
(3a-1)n*n+(3a-2b)n+(b-a-c)=0
=>a=1\3 , b = 1\2 ,c= 1\6
F(x)=(1\3)n*n*n+(1\2)n*n
+(1\6)n