maths induction (1 Viewer)

[blakwidow]

Explain the method of mathematical induction, and use it to prove that

xⁿ - 1 is divisibe by x - 1

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when n = 0,
xⁿ - 1 =0 which is divisible by x-1

Assume true for n=k and prove true for n=k+1
x^k -1 = M(x-1)

x^k+1 - 1 is divisble by x-1

=x^k . x -1

=[M(x-1) + 1]x -1

=(Mx-M +1)x - 1

=Mx²-Mx + x -1

=Mx(x-1) +1(x-1)

=(x-1)(Mx+1) which is divisble by x-1

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a)Is this method correct?
b)book has HINT : x^k+1 -1 = x(x^k -1) + (x-1)

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Another one.

Prove the following by mathematical induction, for integers n>= 1 [greater than or equal to]

If T_n+1 = 2T_n + 1 and T₁ =1, then T_n = 2ⁿ - 1

 

[pLuvia]

The hint is pretty much your answer, since x^k-1=M(x-1) then sub that into the answer they gave you and voila, and your method is fine

 

[blakwidow]

ok i see it now

thanks

should i delete this post now?

 

[watatank]

just leave it up, someone else might find it handy.

 

[blakwidow]

ok thanks

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posted another qn anyone willing to help?

 

[vafa]

if p(n+1)=2p(n)+1,p(1)=1, then p(n)=2^n-1
p(1)=1 true
p(k)=2^k-1
p(k+1)=2^(k+1)-1

p(k+1)=2.(2^k-1)+1=2^(k+1)-1
Hence by the principle of Mathematical Induction, p(n) is true for all values of n(the natural numbers).

 

[blakwidow]

I didn't understand it?

How does this show that P(n) = 2^n-1

And how come they gave the fact that if p(n+1)=2p(n)+1

But you didn't use it??

thanks for the help
bye