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Question

(x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+...n terms =

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Solution

It is easy to observe that
x2y2xy=x+y,
x3y3xy=x2+xy+y2 ,etc.
xnynxy=xn1+xn1+xn2y+....+xyn2+yn1
L.H.S.=1xy[(x2y2)+(x3y3)+.....nterms]=1xy[S1+S2ofaG.Pofnterms]
=1xy[S1+S2]ofaG.Pofnterms=1xy[x2(1xn)1xy2(1yn)1y]
Note it may also be noted that x3+y3x+y=x2xy+y2x5+y5x+y
=x4x3y+x2y2xy3+y4 and xn+ynx+y=xn1xn2y+...xyn2+yn1 where n is 3,5,....i.e.odd.

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