Problems in Greatest Binomial Coefficient

Algebra > Binomial Theorem Examples

7. Find the greatest binomial coefficient in the binomial expansion (3x+2y)12.

Solution:

In the given binomial expansion, the index n is 12, an even number. Therefore from our discussion in Greatest Binomial Coefficients , we see that the greatest binomial coefficient will be nc( n/2 + 1).

Therefore it is 12c(12/2 + 1) = 12c(6+1) = 12c7.

 

8. Find the greatest binomial coefficient in the binomial expansion of
(2x – 4y)11

Solution:

In the given binomial expansion, the index n is 11, an odd number. Therefore from our discussion in Greatest Binomial Coefficients, we see that there are two greatest binomial coefficients. And,they are:

nc(n+1)/2 and nc( n + 3 )/2.

Applying 11 in n:

11c(11+1)/2 = 11c6 and 11c(11+3)/2 = 11c7

9. Find the value of

20C1 + 20C3 + 20C5 + 20C7 + ………….. + 20C19

Solution:

From properties of binomial coefficients , we know that

c0 + c2 + c4 +……….. = c1 + c3 + c5 + …………. = 2n – 1

Since only sum of the odd binomial coefficients is required, we can apply the above formula:

20C1 + 20C3 + 20C5 + 20C7 + ………….. + 20C19 = 220 – 1

10. In the binomial expansion (2x2 – 3/x)15, the term which does not have x is 15c10 . 310 . K, where K is a real number. Find K

Solution.

The term is independent of x ( since it does not have x ). Therefore, we will use general term to find r which is independent of x

Tr + 1 = ncr .xn – r. yr

= 15cr ( 2x2) 15 – r ( - 3 / x) r

= 15cr (2)15 – r ( - 3 ) r (x2) 15 – r (1/x)r

= 15cr (2)15 – r ( - 3 ) r (x30 – 2r)(x)- r

= 15cr (2)15 – r ( - 3 ) r (x 30 – 3r)……………………(3)

Since Tr + 1 is independent of x, equate 30 – 3r to 0

30 – 3r = 0, 3r = 30, r = 10.

Since r = 10, it is the 11th term that does not have x.

Plug in r = 10 in (3) above to find K.

= 15c10 (2)15 – 10 ( - 3 )10

= 15c10 ( 3 )10 (2)5 ……………..(4)

Comparing the value in (4) above with 15c10 .310 . K given in the question, we observe that

K = 25 = 32