## Binomial Theorem Examples

Algebra > Binomial Theorem >solved Examples

5. Find the coefficient of x44 in the binomial expansion (2x3 - 3/)18
Solution:
Remember the detailed discussion in Binomial Coefficient of any power of x discussed earlier above. The following steps must be recollected:
• We use the general term
• We must collect all the powers of x in the given binomial expansion
• We must set the powers of x in step 2 to 44. This gives us the value of r.
• We use r value obtained in the general term to find the coefficient of x44
Tr + 1 = ncr .x n – r . yr
= 18cr . (2x3) 18 – r .(-3/ )r
= 18cr . (2) 18 – r . (-3)r . (x3) 18– r .1/ )r
= 18cr . (2) 18 – r . (-3)r . (x)54 – 3r .(x)-2r
= 18cr . (2) 18 – r . (-3)r . (x)54 – 5r ……………………(1)

Since we must find the coefficient of x44, we must equate the power of x that is
54 – 5r in (1) to 44.
54 – 5r = 44
5r = 10, r = 2.

Therefore it is the third term, T3 that contains the coefficient of x44 in the given binomial expansion.
Now we will write this value of r, that is 2, in (1) to write the coefficient of x44
18c2 . (2)18 – 2 . (-3)2
= 1377.(2)16

6. Find the term independent of x in (2x3 + 3/ )10
Solution:
Remember our discussion of independent term of x in a binomial expansion in formula 8 above earlier. Recall the steps as follows:
1. Use the general term first
2. Collect all the powers of x in the given binomial expansion.
3. Term independent of x means the term does not contain x. For the term to be without x or independent of x, the power of x should be equated to 0, since x0 = 1 and thus x vanishes in the term into 1
Tr + 1 = ncr .x n – r . y r
= 10cr . (2x3) 10 – r . (3/ )r
= 10cr (2)10 – r . (x3)10 – r (3)r. (1/ )r
= 10cr (2)10 – r . (x)30 – 3r . (3)r . (x) -2r/sup>
= 10cr (2)10 – r . (3)r . (x)30 – 5r ………………..(2)

Now set the power of x to 0 to make the term independent of x
30 – 5r = 0
r = 6.

Thus, the term independent of x or not having x is
Tr + 1 = T6 + 1 = T7
Hence, the seventh term T7 does not contain x
Lets write 6 in r in (2) to write this independent term of x
= 10c6 (2)10 –6. (3)6 . (x)30 – 5(6)
= 10c6 (2)10 –6. (3)6 . x0
= 10c6 (2)4. (3)6