Binomial Theorem Examples

Algebra > Binomial Theorem Examples

13. Let us understand the Binomial Theorem concepts discussed above with the following numerous solved examples on each of the concepts and formulas:

1. Find the number of terms in the following binomial expansions:

1. (3x + 4y )10 2. (x – 1/x )17

Solution:

We know the number of terms in a binomial expansion is always one more than one, i.e.,n + 1. In question 1, number of terms is 10 + 1, i.e., 11 and in question 2, the number of terms is 17 + 1, i.e., 18.

2. Use the binomial theorem to expand the following binomial expansion or write all the terms in the following binomial expansion:

1. (2x – 1/x)5

Solution:

Let us use the binomial theorem to write all the terms in the two examples in this question. Remember the binomial theorem? It is:

(x+y)n = nc0. xn +nc1. xn-1. y + nc2 xn-2. y2 + nc2 . xn-3. y3 + ……. + ncr. xn-r. yr + ……. + ncn .yn

Compare the given binomial expansion (2x – 1/x)5 with the standard form (x+y)n.

So we see that we must write in the binomial theorem:
2x for x and – (1/x ) for y.

(2x – 1/x)5 = 5c0 . (2x)5 + 5c1 . (2x)4. y + 5c2 . (2x)3. (y)2 + 5c3 .(2x)2 (y)3 + 5c4 .(2x ) (y)4 + 5c5 .(y)5

Write the following values of the binomial coefficients above
5c0 = 1, 5c1 = 5, 5c2 = 10,5c3 = 10, 5c4 = 5, 5c5 = 1.

= 1. (2)5. (x)5 + 5.(2)4.(x)4.(y) + 10.(2)3.(x)3.(y)2 + 10. (2)2.(x)2.(y)3 + 5.(2x).(y)4 + 1.(y)5

= 32.x5 + 80.x4y + 80 . x3 y2 + 40.x2y3 + 10xy4 + y5

3. Write the tenth term in the binomial expansion (4x31/ )13

solution:

Recall that to find any term is to find r. And to write any term when we know r is to use the general term. And the general term is

Tr + 1 = ncr .x n – r . y r


We need to find the 10th term in the given expansion i.e., T10 = T9 + 1

So r is 9. ( also, recall that the value of r is always one less than the term)

Again, compare the standard form of the binomial expansion with the given form to write x and y in the respective places

On comparing, we see that x = 4x3 and y = – 1/ and n = 13

So, the tenth term is:

T9 + 1 = T10 = 13c9 . (4x3)13 – 9. ( – 1/ )9

= 13c9 .(4)4 (x3)4 (-1)9 ( 1/ )9

{ (-1)9 = -1 and 13c9 = 13!/ 9! 4! = ( 13. 12. 11. 10. 9!) / (9! . 24) = 715}

= - 715 (4)4. (x)12.(x)-18           { (x)12.(x)-18 = x -6 }

= -183040. (x)-6



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