Squares of first n integers

Hello Friends,
I am stuck at proving this problem.
Prove that the sum of the squares of the first n integers is
1/24 (2n) (2n+1) (2n+2):)

Thanks.
Harshal.

Hello, Harshal!

How about an inductive proof?

Quote:

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Prove: .1² + 2² + 3² + ... + n² .= .n(n + 1)(2n + 1)/6

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Verify S(1): .1² .= .1·2·3/6 . → . 1 = 1 ... true


Assume S(k): .1² + 2² + 3² + ... k² .= .k(k + 1)(2k + 1)/6


Add (k + 1)² to both sides:
. . 1² + 2² + 3² + ... + (k + 1)² .= .k(k + 1)(2k + 1)/6 + (k + 1)²

The left side is the left side of S(k + 1).


Factor the right side: .(k + 1) [k(k + 1)/6 + (k + 1)]

. . = .(k + 1) (k² + k + 6k + 6)/6 .= .(k + 1)(k² + 7k + 6)/6

. . = .(k + 1)(k + 2)(2k + 3)/6 .= .(k + 1) ([k + 1] + 1)(2[k + 1] + 1)/6

. . and this is the right side of S(k + 1).


The inductive proof is complete.

We know that the sum of first n natural numbers is

(n(n+1))/2

Now we know again by assuming the sum of the squares of first n natural numbers as Sn

x^3-(x-1)^3=3x^2-3x+1

Put x = 1,2,3,.....,(n-1),n successively, we get:

1^3-0^3 = 3*1^2 - 3*1+1

2^3-1^3 = 3*2^2 - 3*2+1
....................................
....................................
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(n-1)^3 - (n-2)^3 =3*(n-1)^2 - 3*(n-1) +1
n^3 - (n-1)^3 =3*n^2 - 3*n +1
----------------------------------------------------
Add column wise and then simplify it a bit by applying the formula for addition of first n natural numbers to get the answer.


------------------------------

We know that the sum of first n natural numbers is

(n(n+1))/2

Now we know again by assuming the sum of the squares of first n natural numbers as Sn

x^3-(x-1)^3=3x^2-3x+1

Put x = 1,2,3,.....,(n-1),n successively, we get:

1^3-0^3 = 3*1^2 - 3*1+1

2^3-1^3 = 3*2^2 - 3*2+1
....................................
....................................
....................................
(n-1)^3 - (n-2)^3 =3*(n-1)^2 - 3*(n-1) +1
n^3 - (n-1)^3 =3*n^2 - 3*n +1
----------------------------------------------------
Add column wise and then simplify it a bit by applying the formula for addition of first n natural numbers to get the answer.


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Hello, jagabandhu!

I like that method . . . I've used it quite often.
. . Let me try to format it.


We know that the sum of first n natural numbers is: .∑k .= .½n(n + 1)

Now consider the difference of two consecutive cubes:
. . x³ - (x - 1)³ .= .3x² - 3x + 1


Now let x = 1, 2, 3, ..., n and we get:

. . 1³ - 0³ . . . .= . . 3·1² . - . 3·1 + 1
. . 2³ - 1³ . . . .= . . 3·2² . - . 3·2 + 1
. . 3³ - 2³ . . . .= . . 3·3² . - . 3·3 + 1
. . . . . . . . . . . . . . . . . . . . . .. . . . .
. . . . . . . . . . . . . . . . . . . . . .. . . . .
. . . . . . . . . . . . . . . . . . . . . .. . . . .
(n-1)³ - (n-2)³ .= .3(n-1)² - 3(n-1) + 1
. . .n³ - (n-1)³ .= . . 3n² . - . .3n . + 1


Now, add these equations.
Nearly all of the left side cancels out.
On the right, we have three sums.

. . n³ .= .3·∑k² - 3·∑k + ∑1

. . n³ .= .3·∑k² - 3·½n(n+1) + n


Then we have: .3·∑k² .= .n³ - n + 3n(n+1)/2

. . . . . . . . . . . . . . . .= .n(n² - 1) + 3n(n+1)/2

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


Factor: . 3·∑k² .= .n(n + 1)/2 · [(2(n - 1) + 3]

. . . . . . .3·∑k² .= .n(n + 1)(2n + 1)/2


Therefore: .∑k² .= .n(n + 1)(2n + 1)/6