Step-by-step answer:
We are looking at the coefficient of the 22nd term of (x+y)^25.
Following the sequence, first term is x^0y^25, second term is x^1y^24, third term is x^2y^23...and so on, 22nd term is x^21y^4.
The twenty-second term of (x+y)^25 is given by the binomial theorem as
( 25!/(21!4!) ) x^21*y^4
=25*24*23*22/4! x^21y^4
= 12650 x^21 y^4
The coefficient required is therefore 12650, for a binomial with unit valued coefficients.
For other binomials, substitute the values for x and y and expand accordingly.
Question would have been more clearly stated if the actual binomial was given, as commented above.
