# If the numerical coeficients are disregarded the folowing terms apear in the expansion of (x + y)11 State the value of k a)x2 yk             b)xk y11

sciencesolve | Certified Educator

You need to remember binomial formula such that:

`(x+y)^n`  = `sum_(k=0)^n` `nCk*x^(n-k)*y^k`

The problem provides the information that the binomial coefficients are not considered, hence, you need to solve the equation:

`x^2*y^k = x^(11-k)*y^k`

Reducing by `y^k`  yields:

`x^2 = x^(11-k)`

You need to equate the coefficients such that:

`2 = 11 - k =gtk = 11 - 2 =gt k = 9`

Hence, evaluating k if `x^2*y^k`  and n = 11, yields  k = 9.

b) You need to solve for k the equation `x^k*y^11 = x^(11-k)*y^k`

You need to equate the exponents of y such that:

`k = 11 =gt x^(11-k) = x^(11-11) = x^0`

Hence, evaluating the exponent of x if k = 11 and n=11 yields `x^0` .

This is a relatively straightforward problem, as the general formula for any given term when given the power of a simple binomial expansion in the form (ax + by)^n and the term number, r, is:

(nCr)((ax)^(n - r))((by)^r)

We see that (n - r) and r add to n. Hence when you want to find an exponent of one of the terms (x, y) and you know the other, you only need to keep in mind that here the two exponents add to 11.

When wanting to find k in (x^2)(y^k), the value of k is just 11 - 2 = 9.

With (x^k)(y^11), k = 11 - 11 = 0. The term just becomes y^11 (Because any number to the power of 0 is 1).

(NB: Harder problems come when you are asked to find the value of the coefficient of any given term in the expansion. In such cases, further use of the formula is necessary).