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 = 12cr .(3x)12 – r .(-1 / 2x2)
= 12cr .(3)12 – r ( x )12 – r . (– 1 /2) r. 1/x2 ) r
= 12cr . (3)12 – r. (– 1 / 2) r ( x )12 – r. ( x ) – 2r
= 12cr . (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:

= 12c4 . 3 (12 – 4) . ( – 1 /2 )4 . x 12 – 12
= 12c4 . 38 . ( 1 / 16 ). 1 (since x0 = 1)
= 38 . ( 495 / 16 ) (since 12c4 = 495)

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