How is this algebraic step performed?

Question

$$1-\frac{k+2}{(k+1)(k+2)}+\frac1{(k+1)(k+2)}=1-\frac{k+1}{(k+1)(k+2)}$$

Why does the equation hold? I thought $k + 2 + 1$ would be $k + 3$ , but its somehow $k+1$ in the example?

this is regarding induction. I just need to understand the algebraic step.

Answer 1

Don't forget the negative sign.

$$-(k+2)+1 = -k-2+1=-k-1 = -(k+1)$$

Another possible view:

$$-(k+2)-(-1)=-(k+2-1)=-(k+1)$$

Answer 2

It might be easier to think of a version of the problem without fractions cluttering it up: $$1-(k+2)+1.$$ All I've done is gotten rid of the denominators of the second and third terms. Note the parentheses around the second term.

I can rewrite the expression above as $$1-k-2+1;$$ all I'm doing here is expanding out "$-(k+2)$."

Now look at that "$-2+1$" bit. $-2+1=-1$, so we have $$1-(k+2)+1=1-k-1.$$ And finally, this can be rewritten as $$1-(k+1)$$ (essentially factoring out a "$-1$" from the last two terms).

Answer 3

The left side is $$1-(k+2)x+x$$ where $x=1/[(k+1)(k+2)]$. But $$1-(k+2)x+x=1+(-(k+2)+1)x=1+(-k-1)x=1-(k+1)x.$$