numerator = _{[ ]}^{[ ]} n(n+1)(n+2) =[n(n+1)(n+2)(n+3)]/4 (use method of differences

or summation formula)

 

denominator = n(_{[ ]}^{[ ]} n(n+1)) = [n(n+1)(n+2)](n)/3

                                           

 

An = [3(n+3)]/4n= (3/4) (1+ 3/n)

 

as n tends to infinite An =3/4 x=3,y=4 

 

substituting these values we have

 

p^3 + q^3 = 8(r)^3 - 6pqr

 

p^3 + q^3 + (-2r)^3 = 3(p)(q)(-2r)

 

so p+q - 2r = 0   p,r,q are in a.p

 

or p = q = -2r  p+q+4r = 0