﻿ Find the sum of the following:n1u00b2 + 2u00b2 + 3u00b2 + 4u - science mathematics
Find the sum of the following:n1u00b2 + 2u00b2 + 3u00b2 + 4u

## Find the sum of the following:n1u00b2 + 2u00b2 + 3u00b2 + 4u

[From: Mathematics] [author: ] [Date: 01-07] [Hit: ]

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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.

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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)
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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.
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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)
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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: 5-1=4, 14-5=9, 30-14=16, 55-30=25, 2nd differences: 9-4=5, 16-9=7, 25-16=9, 3rd differences: 7-5=2, 9-7=2, gives linearity: (2/3!) = 1/3, so n³/3 is a term.

1-1/3=2/3, 1+4-8/3=7/3, 1+4+9-27/3=15/3, 1+4+9+16-64/3=26/3, ...
1st differences: 7/3-2/3=5/3, 15/3-7/3=8/3, 26/3-15/3=11/3, 2nd differences: 8/3-5/3=1, 11/3-8/3=1, gives linearity: (1/2!) = 1/2, so n²/2 is also a term.

1-1/3-1/2=1/6, 1+4-8/3-4/2=1/3, 1+4+9-27/3-9/2=1/2,
1st differences: 1/3-1/6=1/6, 1/2-1/3=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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ikaug say: xflnnxbx
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sfekr say: mkgosgwc
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emcyu say: ucjumfon
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Otto say: nhlqvfmu
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htmen say: qeyqwomc
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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.

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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+ ... +(n-1)+n
0+2+3+4+ ... +(n-1)+n
0+0+3+4+ ... +(n-1)+n
0+0+0+4+ ... +(n-1)+n

0+0+0+0+ ........ +0+n

Let P(n) be the sum of next (n-1) expressions ,
1
1+2
1+2+3
1+2+3+4

1+2+3+4+ ... +(n-1)

If we add S(n) and P(n) then we get same n expressions ,
1+2+3+4+ ... +(n-1)+n

1+2+3+4+ ... +(n-1)+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 (n-1) expressions , we find that
P(n) = (1/2)1(1+1) + (1/2)2(2+1) + (1/2)3(3+1) + ... + (1/2)(n-1)n
= (1/2)[(2-1)2 + (3-1)3 + (4-1)4 + ... + (n-1)n]
So 2P(n) = 1*2 + 2*3 + 3*4 + ... + (n-1)n .

When we calculate P(n) as the sum of gropus of same terms , we find that
P(n) = 1*(n-1) + 2*(n-2) + 3*(n-3) + ... + (n-1)*1

So
2P(n) + P(n) = 1*[2+(n-1)] + 2*[3+(n-2)] + 3*[4+(n-3)] + ... + (n-1)*[n+1]
= 1*[n+1] + 2*[n+1] + 3*[n+1] + ... + (n-1)*[n+1]
= [1+2+...+(n-1)] * [n+1]
= [(1/2)(n-1)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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Handsome say: Visit www.schoolacers.com for help with any homework problems.
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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.
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mizoo say: = n(n + 1)(2n + 1)/6
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jtzzh say: cksvskzh
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Bryce say: Sum= ∞
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say: The correct answer would be: 69.420
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