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```Applications of Binomial Expression

The binomial theorem is an important topic of the IIT JEE Mathematics. The binomial expansion has got immense applications and is extremely useful in simplifying various lengthy computations.

The binomial expansion for the nth degree polynomial is given by:

If we need to compute (1+x)14 that does not mean that we will have to multiply the term 14 times but rather with the help of binomial theorem it can be computed within seconds.

We Discuss Some of the Major Applications of the Binomial Expression:

1. Divisibility problems:

Let (1 + x)n = 1 + nC1x + nC2x2 +...+ nCnxn.

In any divisibility problem, we have to identify x and n. The number by which division is to be made can be, x, x2 or x3, but the number in the base is always expressed in form of (1+x). This is the most convenient and widely used form of binomial expression.

View the video on binomial theorem

Illustration:

Find the remainder when 7103 is divided by 24.

Solution: Such types of questions are generally asked in the exam and are a bit tricky.

We are required to find the remainder when 7103 is divided by 24.

Firstly, we know that 48 is divisible by 24 which can also be written as 49 is congruent to 1 mod 24. Thus, this can be simplified with the binomial expressions as

7103 = 7(50 - 1)51 = 7(5051 - 51C1 5050 + 51C2 5049 - ... - 1)

= 7(5051 - 51C15050 +...+ 51C5050) - 7 - 18 + 18

= 7(5051 - 51C15050 +...+ 51C5050) - 25 + 18

=> remainder is 18.

Illustration:

If I is the integral part and f is the fractional part of (2 + √3)n then prove that (I + f)(I - f) = 1. Also prove that I is an odd integer.

Solution: Given expression is (2 + √3)n.

I + f where I is an integer and 0, f < 1. We have to show that I is odd and that (I + f) (I - f) = 1

Here note that (2 + √3)n (2 - √3)n = (4 - 3)n = 1

.·.(2 + √3)n(2 - √3)n =1. It is thus required to prove that (2 - √3)n = I-f

But, (2 + √3)n + (2 + √3)n = [2n - C1.2n-1.√3 + C22n-2.(√3)2 - ...]

+ [2n - C1.2n-1.√3 + C22n-2.(√3)2 - ...]

=2[2n - C2.2n-2.3 + C42n-4.32 - ...] = even integer

.·.  Now 0 < (2 - √3) < 1

.·.  0 < (2 - √3)n < 1

.·.  If (2 - √3)n = f' then I + f + f' = Even

Now 0 < f < 1 and 0 < f' < 1

.·.  f + f' = integer

The above equations imply that f + f' = 1 (·.· 0 < f + f' < 2)

.·.  I is odd and f' = I - f

=> (I + f)(I - f) = 1.

Binomial Expression with Non- Positive Exponent

The characteristics discussed hitherto were confined to positive integer n. If n takes any other value, then binomial theorem is written as:

Say, we have to find out (x+a)n n ∉ I+

If a > 1 then it is written as an (a + (x/a))n. This is expanded as

an (1+(x/n))n = an[1+n(x/a) + (n(n-1)/2) (x/a)2 +  (n(n-1)(n-2))/3! (x/a)3 + ......∞]

Since n is not positive integer therefore the series on the right hand side will converge only for |x/a| < 1. Moreover, there are infinite terms in the expansion contrary to the binomial expansion for a positive integer n.

Remarks:

It is important to note here that in this case it is not possible to use the symbol nCr to denote the general term.

The expansions in ascending powers of x are valid only if x is small. In case x is large, i.e. |x| > 1, then it is more convenient to expand it in powers of 1/x.

The following expansions must be remembered as they help in solving questions easily (for |x| <1)

(1 + x)-1 = 1 - x + x2 – x3 + x5 - ….. ∞

(1 - x)-1 = 1 + x + x2 + x3 + x4 + ….. ∞

(1 + x)-2 = 1 - 2x + 3x2 – 4x3 + ….. ∞

(1 - x)-2 = 1 + 2x + 3x2 + 4x3 + ….. ∞

Illustration:  Given positive integers r > 1, n > 2 and the coefficient of (3r)th and (r + 2)th terms in the binomial expansion of (1+x)2n are equal. Then what is the relation between n and r?

Solution: In the expansion of (1+x)2n we have

t3r = 2nC3r-1(x)3r-1

tr+2 = 2nCr+1(x)r+1

Since the binomial coefficients of t3r and tr+2 are equal, so

2nC3r-1 = 2nCr+1

This gives 3r -1 = r + 1 or 2n = (3r -1 + r + 1)

Hence, 2r = 2 or 2n = 4r

Hence, r = 1 or n = 2r

But r >1, so we take the value of n = 2r.

Multinomial Expression:

The generalized form of binomial expansion is called the multinomial expansion. When there are more than two terms the case is considered to be of multinomial expansion.

If n ∈ N, then the general term of the multinomial expansion (x1 + x2 + x3 +...+ xk)n is (n!/a1!a2!...ak!) (x1a1,x2a2,..…xkak ), where a1 + a2 + a3 +......+ ak = n and a < ai < n, i = 1, 2, 3, ...... k and the total number of terms in the expansion is n+k-1Cn-1.

askIITians is an online portal which offers a platform where students are free to post any kind of queries on topics like the application of binomial theorem to find the remainder and the related numerical.

Related Resources

Click here for the Complete Syllabus of IIT JEE Mathematics.

Look into the Solved Papers of Previous Years to get a hint of the kinds of questions asked in the exam.

You can get the knowledge of Useful Books of Mathematics

To read more, Buy study materials of Binomial Theorem comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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