manipulation of subtractionI am trying to solve an induction problem and got stuck at this part.1−n+2(n+2)!+n+1(n+2)!=1−(n+2)−(n+1)(n+2)!Shouldn't it be1−n+2(n+2)!+n+1(n+2)!=1−(n+2)+(n+1)(n+2)!How do you get the left expression to the right expression?

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manipulation of subtractionI am trying to solve an induction problem and got stuck at this part.1−n+2(n+2)!+n+1(n+2)!=1−(n+2)−(n+1)(n+2)!Shouldn&#039;t it be1−n+2(n+2)!+n+1(n+2)!=1−(n+2)+(n+1)(n+2)!How do you get the left expression to the right expression?

ncruuk7ikt · Accepted Answer

Would you feel more comfortable with this?1−n+2(n+2)!+n+1(n+2)!=1+(n+1)−(n+2)(n+2)!If you chose to, you could rewrite it from here to1−[−(n+1)−(n+2)(n+2)!]=1−(n+2)−(n+1)(n+2)!

Alonso Christian · Accepted Answer

Note that−A+BC=(−1)×(A+BC)=(−1)×(AC+BC)=−AC−BC=−A−BC.The minus sign in front of the first fraction does influence the plus sign in front of B of the numerator of the first fraction.

manipulation of subtraction I am trying to solve an

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