Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (kmullis6):

Please help, I'll give fan & medal

OpenStudy (kmullis6):

Directions: Complete the conjecture based on the pattern you observe in the specific cases. Conjecture: The product of a number (n - 1) and the number (n +1) is always equal to _____. \[3\times5=4^{2}-1\] \[4\times6=5^{2}-1\] \[5\times7=5^{2}-1\] \[6\times8=7^{2}-1\] \[7\times9=8^{2}-1\] \[8\times10=9^{2}-1\]

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

it is not clear? what do you think \[9\times 11\]would equal in that pattern?

OpenStudy (kmullis6):

99 so odd number. that's the types of questions i'm answering right now.

OpenStudy (misty1212):

lets look more carefully

OpenStudy (misty1212):

the right hand side of each equal sign is some number squared minus 1 right?

OpenStudy (kmullis6):

Yes

OpenStudy (misty1212):

and that number, the number being squared, is half way between the other two numbers right?

OpenStudy (misty1212):

the number in the middle of \(n-1\) and \(n+1\) is just \(n\)

OpenStudy (misty1212):

so what they are showing you, by example instead of by algebra, is that \[(n-1)(n+1)=n^2-1\]

OpenStudy (kmullis6):

so the answer is n?

OpenStudy (misty1212):

lol no, the answer is \(n^2-1\)

OpenStudy (kmullis6):

oohhhh ok. sorry haha :P thank you (:

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!