practice problems pending

 Q1.


S1.
Method 1:
sum = 0
i = 1, j = 1:1, k = 1:1
Add 1

i = 2, j = 1, k = 1:1
Add 1
i = 2, j = 2, k = 1:2
Add 2

First iteration we added:
1
Second:
 (1+2)
Third:
 (1+2+3)

Nth:
. (1+2+3...+n)
Adding all we get:
1 + (1+2) + (1+2+3) .... + (1+...n)
= 1*n + 2*(n-1) .... 1*n
= (k=1:n)Sigma (k*(n-k+1)) 

= sigma(nk - k^2 + k)
= (n+1)*[n(n+1)]/2 - n(n+1)(2n+1)/6
= n(n+1)(n+2)/6

Method 2:

 reduce it step by step:
Triple sigma 1 = double sigma j = sigma i*(i+1)/2 then separate and do the sum.

Q2. Find the sum of the products of every pair of the first n natural numbers.
S2.
1*2 + 1*3 + 1*4 ... 1*n
          2*3 + 2*4 ... 2*n
............................
                               (n-1)*n

= (k = n:2)Sigma(k * {k*(k-1)/2}
= 1/2[sigma(n^3) - 1 - sigma(n^2) + 1] = 1/2[sigma(n^3) - sigma(n^2)] simplify from here and get the answer.

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