2bornot2b
  • 2bornot2b
Prove that the product of 4 consecutive integers is 1 less than a perfect square.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what have you tried?
anonymous
  • anonymous
oops type-0! :( \[n(n+1)(n+2)(n+3)\]
2bornot2b
  • 2bornot2b
Tried many things in vain

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Kidds stuffs, Let the consecutive integers be \( (n - 1), n, (n + 1), (n + 2) \) Expanding: \[n^4 + 2n^3 - n^2 - 2n= (n^2 + n - 1)^2-1 \]
anonymous
  • anonymous
Nice work ffm!

Looking for something else?

Not the answer you are looking for? Search for more explanations.