Suppose we need to find the binomial coefficient of x^{27} in (x^{2}+ 2x )^{15}

Let us see the method step by step as follows:

1. First, we will have to find the term in which x^{27} 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 x^{27}, 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

(x^{2}+2x )^{15}using the general term.

Now, let us apply all the four steps discussed above to find the binomial coefficient of x^{27}in
( x^{2} + 2x )^{15}. First, write the general term

(You must compare the given binomial expansion ( x^{2} + 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 x^{2} 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 (x^{2} + 2x )^{15} the general term is:

**T _{r + 1} = ^{15}c_{r} . (x^{2}) ^{n – r} . (2x)^{ r}**

Now,

= ^{15}c_{r} . (x ^{2 n – 2 r} ) .( 2^{r} x^{r} )

= ^{15}c_{r} . (x^{ 2 n – 2 r + r} ) . ( 2^{r} )

= ^{15}c_{r} . (x ^{2 n – r} ) . ( 2^{r} )………….. (B)

Since we need x^{27} ,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

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