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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 7

Lesson 7: Series basics challenge

# Sum of n squares (part 1)

What is the sum of the first n squares, 1 + 4 + 9 + 16 + ... + n²? In this video we begin the journey towards finding a formula for this sum. Created by Sal Khan.

## Want to join the conversation?

• - For a linear equation, the difference between successive terms is a constant.
- For a quadratic equation, the difference of the difference between the successive terms is a constant; and
- For a cubic equation, the difference of the difference of the difference of successive terms is a constant.

I took this observation on its face value and understood the video but what I'm wondering is...

Where does this insight come from? Is there a proof to it? Has Sal done a video which explains it in better detail?

Many thanks!
• Here are proofs of those three statements:
Proof for a linear equation of the form L(n) = A*n + B, where A and B are constant coefficients. The difference between successive terms of L(n) can be represented by:
L(n+1) - L(n) = (A*(n+1)+B) - (A*n+B) = A*(n+1) + B - A*n - B = A*(n+1) - A*n = A, which we defined as a constant. So then the difference between successive terms of a linear equation is constant.
Proof for a quadratic equation of the form Q(n) = A*n^2 + B*n + C, where A, B, and C are constant coefficients.
The difference between successive terms can be represented by:
Q(n+1) - Q(n) = (A*(n+1)^2 + B*(n+1) + C) - (A*n^2 + B*n + C) = A*(n^2+2n+1) + B*(n+1) + C - A*n^2 - B*n - C = A*(n^2+2n+1) - An^2 + B*(n+1) - B*n = A*(n^2+2n+1 - n^2) + B*(n+1 - n) = A*(2n+1) + B*1 = A*(2n+1) + B.
Which is a linear function of n. So then Q(n+1) - Q(n) is a linear function of n and so the difference between successive terms of a quadratic series is always a linear function of n. We already proved that the difference between successive terms of a linear function is a constant, so this means that the difference between the successive terms of Q(n+1) - Q(n) is a constant, and so the difference of the difference between successive terms of Q(n) is constant. Q(n) is just any quadratic, so the difference of the difference between the successive terms of a quadratic is always constant. See if you can prove the statement for any cubic equation on your own. It's the same general method as for the linear and quadratics.
• "You'll appreciate this even more when you get into calculus."-Sal,
Was he referring to how the second derivative of x^2 is 2?
• I believe what he was getting at is that the second derivative of a quadratic equation is always a constant, and because his second step in looking at the change between adjacent numbers did not produce a constant change, the formula can't be quadratic, which is why we go on to determine that it must be a cubic function.
• hi Sal, I don't understand why this general function? especially why n^3, I mean I didn't get the connection between the difference of the terms and where comes the function. please explain to me
• Sal figures out that the function that results in the behavior of this series is a third degree function by looking at the differences in the series when n is 0, 1, 2, 3, and 4. If those differences were constant, that would mean that the function to model the behavior of this series would be a linear function. But the differences aren't constant. So he looks at the differences of the differences. If those were constant, which they aren't, that would mean that the function to model the behavior of this series would be a quadratic function. When he looks at the differences of the differences of the differences he does find a constant value. That means this is a cubic function. The general form of a cubic function is An^3 + Bn^2 + Cn^1 + Dn^0 or, written more simply: An^3 + Bn^2 + Cn + D, where A, B, C, and D are real numbers (including zero).
• At sal says the sum of 1^2 + 2^2 + ... is equal to An^3 + Bn^2 + Cn + D.
I understand why he used An^3 + Bn^2 + Cn + D but any chance there's a video which explain that in further details?
• A^3 Bn^2 +Cn^ + D is just the general form of a cubic equation. if he had determined that the series was quadratic he would have used An^2 + Bn + C .. does that help?
• At Sal determines that the function that gives the sum of the first n terms of the series must be cubic because the differences of the difference of the differences is constant. Is there any reason why it couldn't be some flukey quartic or quintic, or is it simply an educated guess?
• Am I correct in saying that Sal finds the differences from the sequence of the partial sums?

If so, is the notation {Sn} = 0, 1, 5, 14, 30... correct to use instead of the gaps and sigma notation Sal used from onwards?
• Ok so i was wondering how would I solve something like ((2i)^2)+i from 1 to 10?

I got 10((5+410)/2). Which evaluates to 2075. I know it is wrong, so how would I solve it?

Ps. This is not a homework question
• Hello people, so I realize that there are quite a few number of concepts introduced in this video, all of which seem to be large in size.

First, is the concept of finding the difference of the difference of something, or the d of the d of the d of the d etc. Now, knowing mathematicians, I presume there is a shorthand for this?

Second is the fact that the difference of a quadratic can be modelled by a cubic one. Is there a proof for this somewhere, which may provide some intuition?

Third, I never knew that a systems of equations could be used to solve something such as this. Where does the terms A, B, and C come from, and why is there coefficients n^3, n^2, and n? What if the formula was only in the form of An^3+Cn, or A^n3+Bn^2, and so on.

Any help would be much appreciated ^.^
• The set up of the problem has to do with whether you are working backwards, graph to equation, or forwards, from equation to graph. Each variable represents the different value that is given by a graph, or in a n equation to transfer to a graph. The reason why the equation is with the A,B, and C terms along with the coefficients n^3, n^2, and n are because in the computing of the problem. With the example of . . .

Area PQRS = the summation of x^2 and x^1 times dx/V = 20 Sumation x^2 and x^1 dx/x = 18.33 hr.

In this equation f(x) is a function of x [b, f(x) =1/V] . Plot x and measure the area under the curve between the vertical lines at x = x^1 and x= x^2.

Pqrs is approx. 36.4 squares and the square LMRO is 100 squares and has an area of o.10 X 500 = 50 (hr.).

Thusly, the integral is 36.4/100 X 50 = 18.2 hr.

Compared with the correct value of 18.33 hr.

The error is very little. Does this response help?
(1 vote)
• How about if you have to find the sum of the same number but it's exponent continuously increases? Like 2/3 + (2/3)^2 + (2/3)^3 + (2/3)^4 + (2/3)^5? Is there a video for something similar to that?
(1 vote)
• So what would the sum from i=0 to n of i^pi be? How does this work with irrational powers?
(1 vote)
• That is an exceedingly advanced problem. I think you'd have to use a Generalized Harmonic for that. I'm not sure, that is probably graduate school level calculus.