How to Find the Sum of Squares from 1 to k?

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Homework Help Overview

The discussion revolves around finding the sum of squares from 1 to k, specifically the expression 1^2 + 2^2 + ... + k^2 and its proposed formula, which is k(k+1)(k+2)/6. Participants are exploring the validity of this formula and discussing methods to derive it.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants suggest proving the formula by induction or evaluating the sum as a polynomial of degree 3. Others question the validity of the proposed formula by testing it with specific values of k.

Discussion Status

The discussion is active, with various methods being proposed to derive the sum of squares. Some participants express confusion about the equations generated from the polynomial approach, while others challenge the correctness of the initial formula. There is no explicit consensus yet, but multiple lines of reasoning are being explored.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the extent of guidance provided. There are indications of differing interpretations of the formula's validity, and some participants are questioning assumptions made in the initial posts.

transgalactic
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[tex] 1^2+2^2+..+k^2=\frac{k(k+1)(k+2)}{6}[/tex]

how to get this general result
??

i know

1+2+..+n=\frac{n(n+1)}{2}

so by that rule the sum should be
[tex] \frac{n(n^2+1)}{2}[/tex]
 
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Two methods come quickly to mind.

First, since you are given that the sum should equal

[tex] \frac{k (k+1)(k+2)}{6}[/tex]

prove the result by induction.

Alternatively, in general the sum

[tex] 1 + 2^r + 3^r + \dots + k^r[/tex]

for integer [tex]r[/tex] is a polynomial of degree [tex]r + 1[/tex]. For [tex]r = 2[/tex]

[tex] 1 + 2^2 + 3^2 + \dots + k^2 = ak^3 + bk^2 + ck + d[/tex]

Evaluate the left side for four different values of [tex]k[/tex] (0, 1, 2, 3) and solve the system of linear equations for the coefficients.
 
transgalactic said:
[tex] 1^2+2^2+..+k^2=\frac{k(k+1)(k+2)}{6}[/tex]

Or you can use the teleschoping series, to prove that.

i.e

[tex]\sum_{i=1}^{k}i^2=\frac{k(k+1)(k+2)}{6}[/tex]
 
statdad said:
Two methods come quickly to mind.

First, since you are given that the sum should equal

[tex] \frac{k (k+1)(k+2)}{6}[/tex]

prove the result by induction.

Alternatively, in general the sum

[tex] 1 + 2^r + 3^r + \dots + k^r[/tex]

for integer [tex]r[/tex] is a polynomial of degree [tex]r + 1[/tex]. For [tex]r = 2[/tex]

[tex] 1 + 2^2 + 3^2 + \dots + k^2 = ak^3 + bk^2 + ck + d[/tex]

Evaluate the left side for four different values of [tex]k[/tex] (0, 1, 2, 3) and solve the system of linear equations for the coefficients.

i can't understand how you get the equations out of you second alternative methos
 
transgalactic said:
[tex] 1^2+2^2+..+k^2=\frac{k(k+1)(k+2)}{6}[/tex]

how to get this general result
You can't, because it's not true. Try it with k=2. 12+22=5, 2*3*4/6=4.
i know

1+2+..+n=\frac{n(n+1)}{2}

so by that rule the sum should be
[tex] \frac{n(n^2+1)}{2}[/tex]
What makes you think that? Hint: It's not valid either.
 
statdad said:
Alternatively, in general the sum

[tex] 1 + 2^r + 3^r + \dots + k^r[/tex]

for integer [tex]r[/tex] is a polynomial of degree [tex]r + 1[/tex]. For [tex]r = 2[/tex]

[tex] 1 + 2^2 + 3^2 + \dots + k^2 = ak^3 + bk^2 + ck + d[/tex]

Evaluate the left side for four different values of [tex]k[/tex] (0, 1, 2, 3) and solve the system of linear equations for the coefficients.
Thanks, I found this very interesting :smile:


Through the method shown by statdad, this sum is actually:

(pending for you to find out)
 
Last edited:
Bingo! :smile: But you shouldn't give answers away. :frown:
 

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