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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).
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OK, note that the part in
blue below is just P(k).
Σj=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 Σ
j=1kj
3 from the left side of the above, and you can also subtract (1/4)x
2(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)(4k
3+12k
2+12k+4)
or...
(k+1)
3=k
3+3k
2+3k+1,
which is an identity.
edit: fixed an omission