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Proposition: 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2) = [n(n+1)(n+2)(n+3)]/4

Step (1): If n=1 then LHS (left hand side) = 6, and RHS = 6

Thus, P1 is true.

Step (2): If Pk is true then

k(k+1)(k+2) = [k(k+1)(k+2)(k+3)]/4

Now,

k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

k(k+1)(k+2) +[k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + (k+1)(k+2)(k+3)

[k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4 ---> Common denominator - 4.

Step (1): If n=1 then LHS (left hand side) = 6, and RHS = 6

Thus, P1 is true.

Step (2): If Pk is true then

k(k+1)(k+2) = [k(k+1)(k+2)(k+3)]/4

Now,

k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

k(k+1)(k+2) +[k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + (k+1)(k+2)(k+3)

[k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4 ---> Common denominator - 4.

__Then...what do i do? any clues? I have not proven my proposition, yet.__
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