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