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Discussion Response Post to: Mathematical Induction (Urgeant, very urgent!!!)

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let s(n)=x²ⁿ-y²ⁿ=(xⁿ+yⁿ)(xⁿ-yⁿ). put n=1 then s(1)=(x+y)(x-y), which is

clearly divisible by (x+y).

assuming s(k) is true s(k)=x^2k-y^2k is divisible by (x+y), then

x^2k-y^2k=(x+y)q for some q...........(1)

s(k+1)=x^2[k+1]-y^2[k+1]=x^2k+2-y^2k+2=x^2k+2-x²y^2k+x²y^2k-y^2k+2=x²[x^2k-y^2k]+y^2k[x²-y²]

[by adding &subtracting x²y^2k]

now, s(k+1)=x²(x+y)q+y^2k(x+y)(x-y) {from (1)}

hence s(k+1)=(x+y)[qx²+(x-y)y^2k] this is divisible by(x+y)

hence proved.

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