# Can someone solve me this step by step

### Can someone solve me this step by step

lim((1*1+2*2+3*3+...+n*n)/(n*n*n))
The solution is 1/3
Guest

### Re: Can someone solve me this step by step

That is better written (1^2+ 2^2+ 3^2+ …+ n^2)/n^3. There is a formula for a sum of squares: 1+4+ 9+ 15+ ….+ n^2= (1/6)n(n+ 1)(2n+ 1)= (1/6)(2n^3+ 3n^2+ n).

So you sum, over n^3 is equal to (1/6)(2n^3+ 3n^2+ n)/n^3= (1/6)(2+ 3/n+ 1/n^2).

What is the limit of that as n goes to infinity.
Guest

### Re: Can someone solve me this step by step

If you are wondering about that formula for "sum of squares", there are several ways to derive it. The simplest is to recognize that the sum of powers, sum i^k for fixed k and I from 0 to n is a polynomial in n to degree k+ 1. In particular, the sum of the first n square, 1+ 4+ 9+ 16+ …+ n^2, is a cubic polynomial is n: An^3+ Bn^2+ Cn+ D. We want to determine A, B, C, and D and we need four equations to do that. When n= 0, An^3+ Bn^2+ Cn+ D= D= 0. When n= 1, An^3+ Bn^2+ Cn+ D= A+ B+ C+ D= 0+1= 1. When n= 2, An^3+ Bn^2+ Cn+ D= 8A+ 4B+ 2D+ D= 0+ 1+ 4= 5. When n= 3, An^3+ Bn^2+ Cn+ D= 27A+ 9B+ 3C+ D= 0+ 1+ 4+ 9= 14.

Since we have immediately that D= 0, we have the three equations A+ B+ C= 1, 8A+ 4B+ 2C= 5, and 27A+ 9B+ 3C= 14 to solve for A, B, and C. Multiply the first equation by 2, to get 2A+ 2B+ 2D= 2, and subtract from the second: 6A+ 2B= 3. Multiply the first equation by 3, to get 3A+ 3B+ 3C= 3, and subtract from the third: 24A+ 6B= 11. Multiply 6A+ 2B= 3 by 3, to get 18A+ 6B= 9, and subtract from 24A+ 6B= 11 to get 6A= 2 so A= 1/3. Then 6A+ 2B= 2+ 2B= 3, 2B= 1 so B= 1/2. Finally A+ B+ C= 1/3+ 1/2+ C= 5/6+ C= 1 so C= 1/6.

We have 1+ 4+ 9+ …+ n^2= (1/3)n^3+ (1/2)n^2+ (1/6)n= (1/6)(2n^3+ 3n^2+ n)= (1/6)n(2n^2+ 3n+ 1)= (1/6)n(n+ 1)(2n+ 1).
Guest