Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Binomial Expansion: (n choose k) + (n choose (k+1))...need to use the factorial form to go through to step and eventually prove the answer... = ((n+1) choose (k+1))...

Calculus1
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

\[\begin{align} \binom nk+\binom n{k+1}&=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k-1)!}\\ &=\frac{1}{n-k}\left[\frac{n!(k+1)}{(k+1)!(n-k-1)!}+\frac{n!(n-k)}{(k+1)!(n-k-1)!}\right]\\ &=\frac{(n+1)!}{(k+1)!(n-k)!}=\binom{n+1}{k+1}. \end{align}\]
Thank you! My algebra with factorials is a little weak..is there any way you could briefly explain how you got to each step? Thanks again!!
The only algebraic manipulation that's worth mentioning is that \(n!=n(n-1)!\). Both the LHS and the RHS on first line are self-explanatory. On the second line, I factored out \((n-k)^{-1}\) from both terms (notice I multiplied the second term by \(n-k\)) to equate denominators (using the idea I just mentioned). Finally, I just expanded and added the numerators on the third line to prove the equality.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question