Thread: Tricky Binomial Theorem Business

so the question is:

find the values of k and n if when you expand (1+x)^n, the coefficient of the (k+1) term is twice that of the kth term and the coefficient of the (k+10) term is twice that of the (k+11) term.

I tried to do this but to no prevail? Any help?

Am I supposed to have 2 equations? Cause i tried that, but I ended doing a whole page of work only to not get the answer.

Originally Posted by mneox

so the question is:

find the values of k and n if when you expand (1+x)^n, the coefficient of the (k+1) term is twice that of the kth term and the coefficient of the (k+10) term is twice that of the (k+11) term.

I tried to do this but to no prevail? Any help?

Am I supposed to have 2 equations? Cause i tried that, but I ended doing a whole page of work only to not get the answer.

The coefficients are going to be the same as those in Pascal's Triangle.

Why not just draw Pascal's Triangle until you find the row that you need?

Originally Posted by Prove It

The coefficients are going to be the same as those in Pascal's Triangle.

Why not just draw Pascal's Triangle until you find the row that you need?

cause the answer says i would need to write out 29 rows, and frankly i'd prefer to do it through a more mathematical method so i can learn for next time.

thanks tohugh

in binomial coefficient of k+1 term is given by
(n,k)C i.e (n!)/[(k!)(n-k)!]
use this
you will get two equations
[one of the equation is obtained by
(n!)/[(k!)(n-k)!]=2(n!)/[(k-1)!(n-k+1)!]
i.e coefficient of k+1 term is twice of coefficient of k term.
solving it we get
2k=n-k+1 or
3k=n+1
other equation can be found similarly]
solving them will give
n=29
k=10