## Independent term of x in (x + y)n:

Let’s think what does a term independent of x in Binomial Theorem mean?

It is the term in which the power of x is 0.

Remember the laws of exponents? **x 0 = 1.**

You know how to find the term in which x27 exists from the discussion in No. 7 above.

You made use of the general term * Tr + 1*, you collected all the powers of x in the given binomial expansion and, you set the simplified collected powers of x to 27.

**To find the independent term of x, what you should do?**

You must set the power of x to 0.

Let us apply all of the steps discussed above in the following example:

Find the term independent of x in *(3x – 1 / 2×2 )12*

**Solution:** we very well understand that to find a term is to find r. And, to find r means to use the general term. Collect all the powers of x and set it to 0 to find r. The general term in the standard form of binomial expansion (x + y)n is

**Tr + 1 = ncr .x n – r . y r .……………..(C)**

Comparing it with the given form *(3x – 1 / 2×2 )12*

We observe that in (C), we must write 3x in x and – 1 / 2×2 in y and 12 for n

**T;r + 1 = 12cr .(3x)12 – r .(-1 / 2×2)**

** T; _{r + 1} = ^{12}c_{r} .(3x)^{12 – r} .(^{-1} / _{2x}^{2})**

**=**^{12}c_{r}.(3)^{12 – r}( x )^{12 – r}. (^{– 1 }/_{2})^{ r}.^{1}/_{x}^{2})^{ r}

**=**^{12}c_{r}. (3)^{12 – r}. (^{– 1}/_{2})^{ r}( x )^{12 – r}. ( x )^{– 2r}

**=**^{12}c_{r}. (3)^{12 – r}. (^{– 1}/_{2})^{ r}. ( x )^{ 12 – 3r}….………….. (D)Since we need a term independent of x, which means the power of x must be 0, we will set the power ** 12 – 3r** of x in (D) above to 0.

**12 – 3r = 0, 3r = 12, r = 4**So, the term not having x or independent of x is

*T r + 1 = T4 + 1 = T5*Now, let us use the general term to write this term independent of x or better substitute ** r = 4** in (D) above to write this independent term:

* = ^{12}c_{4} . 3 ^{(12 – 4)} . ( ^{– 1} /_{2} )^{4} . x ^{12 – 12} *

*=*^{12}c_{4}. 3^{8}. (^{1}/_{16}). 1 (since x^{0}= 1)

*= 3*^{8}. (^{495}/_{16}) (since^{12}c_{4}= 495)