Binomial Theorem Guide

The binomial expressions are connected with +ve or -ve signs; a binomial expression consists of two distinct terms. The binomial theorem provides an easy method to expand and solve a binomial function to its power. The total terms in a binomial expression are given as one plus the number of exponents of the binomial or one plus the binomial power. In the following article, which works as a binomial theorem guide, we will try to solve some problems of binomial expressions. We will learn to expand and find the rth term using the binomial theorem. We will also examine calculations of binomial coefficients.

Overview of Binomial Theorem Guide

Solved Problems

Question 1: In a binomial expression, (y + 1/y)2r. Find the rth term of the binomial.

Solution: The given expression is,

(y + 1/y)2r.

We know, Tr+1 = nCr.an-r.br.

Implying, Tr = 2rCr – 1.(y)2r-r+1. (1/y)r-1,

Tr = [(2r)! / (r – 1)!(r + 1)!].yr+1-r+1

Tr = [(2r)! / (r – 1)!(r + 1)!]. y2.

Question 2: Expand the following binomial expression

(1 – z + z2)4

Solution: let us consider 1 – z = x, in the expression,

(1 – z + z2 )4 = (x + z2 )4,

= 4C0 x4 (z2)0 + 4C1 x3(z 2) + 4C2 x2 (z2 )2 + 4C3 x (z2 )3 + 4C4 (z2)4,

= x4 + 4×3 z2 + 6y2z4 + 4x.z6 + z8,

= (1 – z)4 + 4.z2.(1 – z)3 + 6.z4.(1 – z)2 + 4.z6.(1 – z) + z8,

= 1 – 4.z + 10.z2 – 16.z3 + 19.z4 – 16.z5 + 10.z6 – 4.z7 + z8.

Question 3: Find the value of binomial coefficient at y11 for the following binomial expression

[y3 – 2/y2]12

Solution: Let us consider that the general term of the expression (r + 1)th, contain y11

We know, Tr+1 = nCr.an-r.br.

Implying here, Tr+1 = 12Cr (y3 )12-r.(-2/y2)r

= 12Cry36-3r-2r. (-1)r.2r,

= 12Cr.(-1)r .2r.y36– 5r

Now, for this binomial expression to contain y11, we note that,

36 – 5r = 11,

Or, r = 5.

Therefore, the coefficient of y11 is given by,

12C5.(-1)5.25 =  -25344

Therefore, the coefficient of y11 is  -25344.

Question 4: In a binomial expression, (2 + 3z)9, find the numerically greater term, z = 3/2.

Solution: The given binomial expression is,

(2 + 3z),

= 29.[1 + 3z/2]9

Now, Tr+1 /Tr = 29.[9Cr(3z/2)r] / 29[ 9Cr-1.(3z/2)r-1]

9Cr/9Cr-1.|3z/2|,

9! / [(r)!.(9-r)!].[(r-1)!.(10-r)!/9]. |3z/2|,

(10-r)/r . |3z/2|,

(10-r)/r . |9/4| (z = 3/2),

Therefore, Tr+1/Tr ≥ 1,

(90 – 9r)/4r ≥ 1,

90 – 9r ≥ 4r,

r ≤ 90/13,

Dividing the above expression, we know that the maximum value of r is 6,

This gives that the greatest term of the expression is T7 = Tr+1,

The calculation for T7,

T7 = 29[ 9C6.(3z/2)6]

T7 = 29.9C6(9/4)6,

T7 = 29[(9).(8).(7) / (3)(2)(1)] (312/212).

T7 = (7).313 / 2.

Question 5: In a binomial expression, 24a + 4 – 15a – 16. Prove that the binomial is divisible by 225. The value of a is a natural number.

Solution: We have a binomial expression as,

24a+4 -15a -16,

 24.(a+ 1)-15a -16,

16a + 1-15a-16,

(1 + 15)a+1-15a-16.

Expanding, 

a+1C0 .150 + a + 1C1 .151 + a + 1C2 .152 + a + 1C3 .153 + … + a+1Ca+1 (15)a+1 -15a -16,

= 1 + (a + 1).15 + a+1C2.152 + a+1C3.153 + … + a + 1Ca+1.(15)a+1 -15a -16,

=  1 + 15a + 15 + a + 1C2 .152 + a + 1C3 .153 + … + a + 1Ca + 1.(15)a + 1 – 15n – 16,

= 152 [a+1C2 + a+1C3 15 +…so on],

Therefore, the given binomial is divisible by 225.

Question 6: for the following binomial expression, find its middle term.

(a/y + y/b)9

Solution: The given binomial is,

(a/y + y/b)9

Here, we can see that the power of the binomial is an odd number, therefore, it will have two middle terms (5th and 6th terms).

The two middle terms of the binomials will be,

T5 = 9C4 (a/y)5.(y/b)4,

= 9C4. a/y

= 125a/y.

For T6,

T6 = 9C5 (a/y)4.(y/b)5

= 9C5.y/b

= 126y/b.

Conclusion

According to the binomial theorem guide in these notes, binomials are defined as functions with two such terms connected by a positive or negative algebraic relation. The examples provided give us the calculations for the rth term, coefficient of rth term, and divisibility of a binomial expression. We also understand how to expand a binomial expression from the given problems. The solved examples offer a step-by-step guide to understanding how a problem with binomial expression is solved.