Need for 'Pi'

'Pi' the miracle number
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I was asked( long back may be 4 years) to write which known formula provides maximum number of digits to the value of 'pi'. Since my tutor is soft enough, I bravely (rare thing from me :) ) wrote in the paper that "I do not know the formula, but using Ramanujan's formula( which is what he expected, he is a fan/ devotee of great Ramanujan) we can calculate up to some thousands and lakhs of digits, at the same time, in my opinion, the universe radius is in the order of 10 to the power 15 or 20, even if we want to calculate the volume of the universe, it is enough to find around 60 to 70 digits accuracy, I do not know the motivation of finding so many digits for pi, may be this is due to mathematicians' dream/ ego to find out this miracle (maximum number of digits)" !!

Then he explained one of the applications is to test the computation complexity using 'pi' related findings !! 

Now when I saw this article, I remember those days !!

http://en.wikipedia.org/wiki/Pi#Motivations_for_computing_.CF.80

I should have read this article before writing in the paper, but what to do I thought about this after when I read the question !! 

Sum of cube of first n natural numbers is the square of the sum of first n natural numbers !!

Proof for (Simple Try) !!
^{Sum of cube of first n natural numbers is the square of the sum of first n natural numbers !!}

1³+2³ + 3³+ … +n³ = (1 + 2 + 3+...+n) ² 

As we know the simple formula
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(a + b) ² = a² + 2*ab + b ²

(a + b + c) ² = a²+ b²+ c ²+ 2 (ab + ac 
                                                    + bc)

(a + b + c + d) ²= a²+ b²+c²+ d²+ 
                                                     2(ab + ac + ad 
                                                             + bc + bd 
                                                                     + cd )

The above algebric formula has some pattern ( marked as bold) .. 2 ( a (b+c+d) + b (c+d) + cd))

Now let us look into (1 + 2 … + n) ²

(1 + 2 … + n) ^{2   =} 
             
           1²+ 2² + 3² .. + n² + 2*{1(2+3+4..n) + 2 (3+4+...+n) + ... (n-1)(n)}

             

 (1 + 2 … + n) ²= 
                             1²+ 2² + 3² .. + n² +  2*{1*2 + 1*3 + 1*4 + …+ 1*n

                                           + 2*3 + 2*4+ … + 2*n

                                                    + 3*4+…. + 3*n

                                                             +…..+ (n-1) (n)}

Consider the violet terms, this can be written as

2 * 1 + 3 (1+2) + 4 (1+2+3) + …+ (n)(1+2+…(n-1))

using some of first n terms formula ... ( n * (n+1) / 2)) 

The above terms can be rewritten as

2 + 3(2 * 3/2) + 4(3*4/2) + 5(4*5/2) + …. +n((n-1)n/2))

After simplifying, This can be written as

(1 + 2 … + n) ²= 1²+ 2² + 3² … + n² + 2 {2 + 3(2*3/2) + 4 (3*4/3) + ….. n ((n-1) (n/2)}

                          = 1²+ 2² + 3² … + n² + {1*2²+2*3² + 3*4²….+ (n-1)n²}

                          = 1²+ 2² + 3² … + n² + 2^{2 +} 2*3² + 3*4² + ... + (n-1)n² 

upon rearranging the terms

                           = 1²+ (2² +  2²) + (3² + 2*3²) + … + ((n-1)n² + n²) 

                            = 1²+ (2 * 2² ) + ( 3 * 3²) + … + (n* n²) 

                            =1³+ 2 ³+ …+ n ³  

                

(1 + 2 + 3 … +n) ² = 1³+ 2 ³+ … +n ³          

                                              

Change LHS to RHS and vice –versa

1³+ 2 ³+ … +n³    =    (1 + 2 + 3 … +n) ²

FROM the induction ( 1 + 2 + 3 … n ) = ( n (n+1) / 2 ) the above equation can be written as

1³+ 2 ³+ … n³     =    (n (n+1)/ 2) ²