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

[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

 

[CRGreathouse]

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?

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

 

[Little ant]

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

 

[Little ant]

sorry i don't read well the question.

 

[Little ant]

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

 

[Mentallic]

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

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:

 

[Helios]

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...

 

[DDTHAI]

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...

-----------------------------------------------
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!

 

[DDTHAI]

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

 

[DDTHAI]

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

Once again, thank you all for helping.

DDTHAI

 

[Gokul43201]

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

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

 

[awkward]

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
$$\sum_{k=1}^n k^2$$.