I stumbled upon a problem in physics where I had to find this sum. Please help me figure it out.

answers:
Dixon say: I you have seen any visual explanation for why a pyramid or cone has a volume of 1/3 the equivalent prism, and/or seen the "staircase" visual explanation for why 1 + 2 + ... + n = n(n + 1)/2, then you should be able to follow this 1 minute video for the sum of incrementing squares.
https://www.youtube.com/watch?v=NcSQPqF...

Ian H say: A pattern can be detected by multiplying partial sums by 6.
6(1 + 4) = 30 = 2*3*5
6(1 + 4 + 9) = 84 = 3*4*7
6(1 + 4 + 9 + 16) = 180 = 4*5*9
6(1 + 4 + 9 + 16 + 25) = 330 = 5*6*11
6S(n) = n(n + 1) (2n + 1)

jibz say: Noting that
k² = (0² + 1²) + (1² + 2²) + ... + ((j–1)² + j²) + ... + ((k–2)² + (k–1)²) + ((k–1)² + k²)
and that
(j–1)² + j² = 2j – 1,
we have
Σ k² [k=1..n]
= Σ Σ (j–1)² + j² [j=1..k] [k=1..n]
= Σ Σ 2j – 1 [j=1..k] [k=1..n]
= Σ row k in the sum below [k=1..n]
=
[row 1]: 1 +
[row 2]: 1 + 3 +
[row 3]: 1 + 3 + 5 +
·
·
·
[row k]: 1 + 3 + 5 + . . . + 2k–1 +
·
·
·
[row n]: 1 + 3 + 5 + . . . + 2k–1 + . . . + 2n–1
[col. . . .1. . 2. . 3. . . . . . . k . . . . . . . . . n]
= Σ col. k in the sum above [k=1..n]
= Σ (n–k+1)(2k–1) [k=1..n]
= Σ 2k² + (2n+3)·k – (n+1) [k=1..n].
Thus
Σ k² [k=1..n] = 2·Σ k² [k=1..n] + (2n+3)·Σ k [k=1..n] – (n+1)·Σ 1 [k=1..n],
which we can solve for Σ k² : [k=1..n] in terms of sums expressible in closed form:
Σ k² [k=1..n]
= (1/3)((2n+3)·Σ k [k=1..n] – (n+1)·Σ 1 [k=1..n])
= (1/3)((2n+3)(n+1)(n/2) – (n+1)(n))
= (n)(n+1)(2n+1)/6.

Krishnamurthy say: 1² + 2² + 3² + 4² +......+ n²
sum_(k=1)^n k^2 = 1/6 n (n + 1) (2 n + 1)
1 + 2 + 3 + 4 +......+ n
sum_(k=1)^n k = 1/2 n (n + 1)

Φ² = Φ+1 say: 1, 1+4=5, 1+4+9=14, 1+4+9+16=30, 1+4+9+16+25=55, ...
1st differences: 51=4, 145=9, 3014=16, 5530=25, 2nd differences: 94=5, 169=7, 2516=9, 3rd differences: 75=2, 97=2, gives linearity: (2/3!) = 1/3, so n³/3 is a term.
11/3=2/3, 1+48/3=7/3, 1+4+927/3=15/3, 1+4+9+1664/3=26/3, ...
1st differences: 7/32/3=5/3, 15/37/3=8/3, 26/315/3=11/3, 2nd differences: 8/35/3=1, 11/38/3=1, gives linearity: (1/2!) = 1/2, so n²/2 is also a term.
11/31/2=1/6, 1+48/34/2=1/3, 1+4+927/39/2=1/2,
1st differences: 1/31/6=1/6, 1/21/3=1/6, gives linearity: ((1/6)/1!) = 1/6, so n/6 is also a term.
The sum of squares from 1² to n² is then ⅓n³ + ½n² + ⅙n, which factorises to ⅙(2n³ + 3n² + n) to ⅙n(n+1)(2n+1).

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Bradley say: unrgoycg

Dixon say: I you have seen any visual explanation for why a pyramid or cone has a volume of 1/3 the equivalent prism, and/or seen the "staircase" visual explanation for why 1 + 2 + ... + n = n(n + 1)/2, then you should be able to follow this 1 minute video for the sum of incrementing squares.
https://www.youtube.com/watch?v=NcSQPqF...

atsuo say: Assume that you know 1 + 2 + 3 + 4 + ... + n = (1/2)n(n+1) and
you do not know 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 = (1/6)n(n+1)(2n+1) .
Let S(n) = 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 .
It can be represented as the sum of next n expressions ,
1+2+3+4+ ... +(n1)+n
0+2+3+4+ ... +(n1)+n
0+0+3+4+ ... +(n1)+n
0+0+0+4+ ... +(n1)+n
：
0+0+0+0+ ........ +0+n
Let P(n) be the sum of next (n1) expressions ,
1
1+2
1+2+3
1+2+3+4
：
1+2+3+4+ ... +(n1)
If we add S(n) and P(n) then we get same n expressions ,
1+2+3+4+ ... +(n1)+n
：
：
1+2+3+4+ ... +(n1)+n
So S(n) + P(n) = [(1/2)n(n+1)] * n = (1/2)(n^3 + n^2) . Therefore
S(n) = (1/2)(n^3 + n^2)  P(n) .
When we calculate P(n) as the sum of (n1) expressions , we find that
P(n) = (1/2)1(1+1) + (1/2)2(2+1) + (1/2)3(3+1) + ... + (1/2)(n1)n
= (1/2)[(21)2 + (31)3 + (41)4 + ... + (n1)n]
So 2P(n) = 1*2 + 2*3 + 3*4 + ... + (n1)n .
When we calculate P(n) as the sum of gropus of same terms , we find that
P(n) = 1*(n1) + 2*(n2) + 3*(n3) + ... + (n1)*1
So
2P(n) + P(n) = 1*[2+(n1)] + 2*[3+(n2)] + 3*[4+(n3)] + ... + (n1)*[n+1]
= 1*[n+1] + 2*[n+1] + 3*[n+1] + ... + (n1)*[n+1]
= [1+2+...+(n1)] * [n+1]
= [(1/2)(n1)n] * [n+1]
= (1/2)(n^3  n)
3P(n) = (1/2)(n^3  n)
P(n) = (1/6)(n^3  n)
Therefore
S(n) = (1/2)(n^3 + n^2)  P(n)
= (1/2)(n^3 + n^2)  (1/6)(n^3  n)
= (1/6)(3n^3 + 3n^2)  (1/6)(n^3  n)
= (1/6)(2n^3 + 3n^2 + n)
= (1/6)n(2n^2 + 3n + 1)
= (1/6)n(n+1)(2n+1)
So 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 = (1/6)n(n+1)(2n+1) is proved .

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john say: As long as it's infinite you cannot produce anymore result than you have. You have to finalize that series with a result like '=10'. Then you have a finite solution and can remove 'n' or the infinite part.
If you want to reduce it infinitely use partial derivatives.
But as a infinite yeven partials will not produce a result you understand.

mizoo say: = n(n + 1)(2n + 1)/6

jtzzh say: cksvskzh

Bryce say: Sum= ∞

say: The correct answer would be: 69.420
