IIT JEE , AIEEE Forum - Mathematics , Algebra - Mathematical Induction (Urgeant, very urgent!!!)

Ashish

Please somebody answer this as soon as possible ! It's absolutely urgent!!!

Prove using mathematical university, x2n - y2n ( 2n is the power) is divisible by x+y.

I shall be very, very, very grateful to whoever answers the question.

Thank you.

ankurgupta91

P(n):x^2n-y^2n
let us assume P(k) is true
i.e x^2k-y^2k is divisible by x+y
nw P(k+1): x^2(k+1)-y^2(k+1)
=x^2k.x^2 -y^2k.y^2
=x^2(x^2k-y^2k)+x^2.y^2k-y^2k+2
=x(x^2k-y^2k)+y^2k(x^2-y^2)
=x(x^2k-y^2k)+y^2k(x-y)(x+y)
as x^2k-y^2k is divisble by (x+y)
thus , nw P(k+1) is also true
thus by principle of mathematical induction P(n) is true

bhargavi

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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