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.
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