Question

1^2+2^2+3^2+...........(n(n+1))/2)^2
Use this formula to show that An= (n^2+2n+1)/4n^2

Use the same procedure to
1^2+2^2+3^2+...+n^2=n(n+1)(2n+1))/6

Answer 1

This is a lot easier to do with pictures, but I'll try to show you an
algebra approach to this formula. It makes use of this cubing pattern:

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

That is, to cube one more than a number x, first cube x, then triple
x^2, then triple x, and then add these to 1. Okay?

1^3 = (0 + 1)^3 = 0^3 + 3 ( 0^2 ) + 3 (0) + 1
2^3 = (1 + 1)^3 = 1^3 + 3 ( 1^2 ) + 3 (1) + 1
3^3 = (2 + 1)^3 = 2^3 + 3 ( 2^2 ) + 3 (2) + 1
4^3 = (3 + 1)^3 = 3^3 + 3 ( 3^2 ) + 3 (3) + 1
etc.
n^3 = (n-1 + 1)^3 = (n-1)^3 + 3 (n-1)^2 + 3 (n-1) + 1

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

now add all these up in columns. The left side is the sum of the
cubes from 1 to n+1:

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

The first column on the right is also the sum of cubes but starting at
0 and ending at n:

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

The next column on the right has 3 times the sum of the squares from
0^2 to n^2

The next column has 3 times the sum of the integers from 0 to n.

The last column has n+1 ones. [not n ones...n+1 ones!]

1. All of the cubes cancel except for (n+1)^3

2. The sum of the squares is what we are looking for ... call this S

3. The sum of the integers 1+2+3+...+n = n(n+1)/2

4. The sum of n+1 ones is just n+1.

Now combine these:

(n+1)^3 = 3S + 3[ n(n+1)/2 ] + n + 1

Bring everything together:

n^3 + 3n^2 + 3n + 1 - 3n^2/2 - 3n/2 - n - 1 = 3S

Simplify:

n^3 + 3n^2/2 + n/2 = 3S

Multiply through by 2 to clear fractions:

2n^3 + 3n^2 + n = 6S

Factor:

n ( 2n^2 + 3n + 1) = 6S

Factor again:

n ( 2n+1) (n+1) = 6S

Rearrange and solve for S = 1^2 + 2^2 + ... + n^2:

S = n(n+1)(2n+1)/6 !

I hope that helps.