Binomial coefficient of any power of x

7. Binomial Coefficient of any power of x in the binomial expansion:

Suppose we need to find the binomial coefficient of x²⁷ in (x²+ 2x )¹⁵
Let us see the method step by step as follows:

1. First, we will have to find the term in which x²⁷ occurs. To find a term implies finding r. So, we will now find r.

2. To find r, we will need the general term. Remember the general term? It is:

3. In x²⁷, since the power of x is 27, we should notice that in the general term having x^{n – r} ,we should set 27 in n – r.But not so early. Why? Notice the next step 4

4. First collect all the powers of x in the given binomial expansion
(x²+2x )¹⁵using the general term.

Now, let us apply all the four steps discussed above to find the binomial coefficient of x²⁷in ( x² + 2x )¹⁵. First, write the general term

T_{r + 1} = ⁿc_r .x^{n – r} . y^r .………….. (A)

(You must compare the given binomial expansion ( x² + 2x )¹⁵ with the standard form ( x + y )ⁿ in order to write x and y in the general term in (A) above).

From comparison, we see that x² in the given form stands for x in the standard form, while 2x in the given form stands for y in the standard form. In short, x = x ² and y = 2x Therefore, in the given binomial expansion (x² + 2x )¹⁵ the general term is:

T_{r + 1} = ¹⁵c_r . (x²) ^{n – r} . (2x) ^r

Now,
   = ¹⁵c_r . (x ^{2 n – 2 r} ) .( 2^r x^r )
   = ¹⁵c_r . (x ^{2 n – 2 r + r} ) . ( 2^r )
   = ¹⁵c_r . (x ^{2 n – r} ) . ( 2^r )………….. (B)

Since we need x²⁷ ,we observe that we must set 27 in 2n – r in (B) above. 2n – r = 27.

Substitute n = 15,

2(15) – r = 27

r = 3.

Since r = 3, T_{r + 1} = T _{3 + 1} = T4

So, it is the fourth term that contains x ²⁷

Apply r = 3 in (B) above, to write the binomial coefficient of x ²⁷. It is