Find Closed-Form Expression for Summation of n^2

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To find a closed-form expression for the summation Q_N = ∑_{z=1}^N z^2, the solution is N(N+1)(2N+1)/6. The method involves using the identity (k+1)^3 - k^3 = 3k^2 + 3k + 1 and summing it from k=1 to N, leading to a telescopic sum that simplifies to (N+1)^3 - 1. This approach provides an equation for Q_N, confirming the closed-form expression. Additionally, it is noted that the sums of squares relate to the equation form (n^3 - n)/24, with only three integers in N where the sum of squares is also a perfect square. Understanding these relationships is key to solving the problem efficiently.
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Hello,
could anyone give me a hint on how to find a closed-form expression for the following summation:

Q_N = \sum_{z=1}^N z^2

where z are positive integers from 1 to N.
Thanks.
 
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The solution is N(N+1)(2N+1)/6. The trick is simple: take the equations: (k+1)^3-k^3=3k^2+3k+1, sum them up from k=1 to N.
 
csopi said:
The solution is N(N+1)(2N+1)/6. The trick is simple: take the equations: (k+1)^3-k^3=3k^2+3k+1, sum them up from k=1 to N.

Uhm...you mean:

\sum_{k=1}^N \left( (k+1)^3-k^3 \right) = \sum_{k=1}^N \left( 3k^2+3k+1\right)

But how can this help, if we don't know the value of the leftmost summation?
Evaluating the rightmost term also requires evaluating a summation of k2, which is what we are trying to find :confused:
 
You do know, it's a telescopic sum and equals (N+1)^3-1. After that, you get an equation for Q_N
 
csopi said:
You do know, it's a telescopic sum and equals (N+1)^3-1.

Ups! that's true...I should have noticed it immediately. Thanks.
 
mnb96 said:
Hello,
could anyone give me a hint on how to find a closed-form expression for the following summation:

Q_N = \sum_{z=1}^N z^2

where z are positive integers from 1 to N.
Thanks.

Also note the following equation form: (n^3 - n)/24. The sums of squares are a special case of this equation form, and incidentally, there are only 3 integers in N | the sum of squares is also square. This holds for 0, 1, 24 and no other.

- RF
 
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