Algebra > Binomial Theorem > Independent Term of x
8. Independent term of x in (x + y)n: First of all, think what does a term independent of x in Binomial Theorem mean? Got it? Found the clue? Yes, 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. Great! Now, to find independent term of x, guess what you should do? You must set the power of x to 0. Great, you guessed it right! Let us apply all of the steps discussed above in the following example:
Find the term independent of x in (3x – 1 / 2x2 )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 / 2x2)12 We observe that in (C), we must write 3x in x and - 1 / 2x2 in y and 12 for n  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)