# Formula for 1^2 + 2 ^2 + +n^2?

1. Sep 14, 2010

### DDTHAI

Formula for 1^2 + 2 ^2 +.... +n^2?

Hello,

I know how to figure out the formula for 1 +2 + 3 + ... + n (i.e., (1+n)n/2 ). But what is the formula for 1^2 + 2 ^2 + 3^2 + .... + n^2? How does one formulate it?

Thanks,
DDTHAI

2. Sep 14, 2010

### CRGreathouse

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

It's a cubic equation with rational coeffcients. You can find these by fitting the curve to the first four points.

3. Sep 14, 2010

### Little ant

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

1^2 + 2^2 + 3^2 + ... + 2^n = 2^(n-1)

4. Sep 14, 2010

### Little ant

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

sorry i don't read well the question.

5. Sep 14, 2010

### Little ant

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

we know than (n-1)^3 = n^3 - 3n^2 + 3n -1. so

0^3 = 1^3 - 3.1^2 + 3.1 - 1
1^3 = 2^3 - 3.2^2 + 3.2 - 1
2^3 = 3^3 - 3.3^2 + 3.3 - 1
.
.
.
(n-2)^3 = (n-1)^3 - 3(n-1)^2 + 3(n-1) -1
(n-1)^3 = n^3 - 3n^2 + 3n - 1

Then, if we plus all, we get
n n n
0^3 = n^3 - 3E (i)^2 + 3E n - E 1
i=1 i=1 i=1

n n n
=>3E (i)^2 = n^3 + 3E n - E 1
i=1 i=1 i=1

n
=> 3E (i)^2 = n^3 + 3(n+1)n - n
i=1 2

After simplify you get

n
E (i)^2 = n(n+1)(2n+1)
i=1 6

*you must know than
n
E (i) = 1 + 2 `+ 3 + 4 + ... + n = n(n+1)
i=1 2

6. Sep 14, 2010

### Mentallic

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

This makes absolutely no sense. I mean, look at it for a second... firstly you failed to notice the pattern correctly since 2^n means 2^1+2^2+2^3 instead of what is shown. And secondly, how can that all equal 2^(n-1) when on the left side of the equation, you already have 2^n which means the term just before the last is 2^(n-1)?

Even though it was probably a typo or something, I just wanted to be annoying :tongue:

7. Sep 14, 2010

### Helios

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

See the book "ascent to orbit", A.C.Clarke's autobiography, chap 24. He list sums of powers up to 8.

Sum n^2 = ( n / 6 )( n + 1 )( 2n + 2 )

Sum n^3 = ( n^2 / 4 )( n + 1 )( n + 1 ) = [ sum n ]^2

etc...

8. Sep 14, 2010

### DDTHAI

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

-----------------------------------------------
with n = 5, the series 1^2 + 2^2 + 3^2 + .... + n^2 = 1+4+9+16+25 = 55

Sum n^2 = ( n / 6 )( n + 1 )( 2n + 2 ) = (5 / 6) * 6 * (2*5 + 2) = 5 * 12 = 60!

Last edited: Sep 14, 2010
9. Sep 14, 2010

### DDTHAI

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

Thanks Helios,

I believe you tried to write
Sum n^2 = ( n / 6 )( n + 1 )( 2n + 1 )

That would be right for all n's that I have tried so far! But how do prove that the formula is right?

Thanks you all for trying to help
DDTHAI

10. Sep 14, 2010

### DDTHAI

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

Thanks Little Ant - I now understand your approach for coming up with the formula!

Once again, thank you all for helping.

DDTHAI

11. Sep 14, 2010

### Gokul43201

Staff Emeritus
Re: Formula for 1^2 + 2 ^2 +.... +n^2?

That is the correct formula. It can be proved by induction.

12. Sep 14, 2010

### awkward

Re: Formula for 1^2 + 2 ^2 +.... +n^2?

Here is one method to derive the formula for the sum of squares which can be extended to other integer powers.

Observe that $$(k+1)^3 - k^3 = 3k^2 + 3k + 1$$ ... (*)

Now sum up (*) for $$k = 1, 2, \dots , n$$.

The sum on the left hand side telescopes, leaving $$(n+1)^3 -1$$.

The sum on the right hand side is
$$3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n k + n$$.

Now equate these two expressions for the sum, apply the formula you already know for
$$\sum_{k=1}^n k$$
and solve for
$$\sum_{k=1}^n k^2$$.

Last edited: Sep 14, 2010