Thread: Proof by Induction View Single Post
Emeritus
PF Gold
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Proof by Induction

 But I can seem to get this part: Assuming P(k) is true, you can cancel it from the left hand side if you also cancel the right side of P(k) from the right side of P(k+1). You should be left with an identity, which proves that P(k) implies P(k+1).
OK, note that the part in blue below is just P(k).

&Sigma;j=1kj3+(k+1)3=(1/4)(k+1)2(k+2)2

What you need to do is expand the right hand side of both P(k) and of P(k+1). Then, you can subtract &Sigma;j=1kj3 from the left side of the above, and you can also subtract (1/4)x2(x+1)2 from the right side of the above (since those two quantities are equal).

What you will be left with is:

(k+1)3=(1/4)(4k3+12k2+12k+4)

or...

(k+1)3=k3+3k2+3k+1,

which is an identity.

edit: fixed an omission