Introduction To Binomial Theorem and JEE Binomial Theorem Previous Year Questions

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Binomial Theorem 

Binomial Theorem deals with the expansion of the powers of a binomial. This article consists of the concept of the Binomial Theorem and JEE Binomial Theorem Previous Year Questions in detail. The students who are facing the JEE examination can refer to this and get benefitted from it.

Statement: It states that for any positive integer n, the sum of two numbers (say x and y) can be expressed as the sum of (n + 1) terms in the form of 

(x + y)n = nC0xny0 + nC1xn-1y1 + nC2xn-2y2 + ….. nCn-1x1yn-1 + nCnxoyn

Here, n ≥ 0 and nCk is a positive integer called the binomial coefficient.This formula is the binomial formula. Binomial Theorem can be denoted using the summation notation as follows:

Example:

(x + y)0 = 1

(x + y)1 = x + y

(x + y)2 = x2 + 2xy + y2

(x + y)3 = x3 + 3x2y + 3xy2 + y3

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

(x + y)6 = x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6

(x + y)5 = x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

Features of the above example are given below:

1] It consists of (n + 1) terms in the expansion of (x + y)n.

2] n is the degree of each term.

3] The powers of x start with n and then decrease to 0.

4] The powers of y start with 0 and then increase to n.

5] It has symmetric coefficients.

One of the easy ways to find the coefficients of the expansion is Pascal’s Triangle.

It is named after French mathematician Blaise Pascal. In this triangle, each row begins and ends with 1. Every number in the triangle is the sum of the two numbers that are present immediately above it.

JEE Binomial Theorem Previous Year Questions

A subject that needs conceptual understanding is mathematics. One of the important concepts in JEE is Binomial Theorem. About two to three questions can be asked from the JEE Binomial Theorem Previous Year Questions. Every question of Binomial Theorem along with the detailed solution is discussed here. 

Question 1: The value of (√2 + 1)6 + (√2 − 1)6 will be 

Solution:

(x + a)n + (x − a)= 2 [xn+ nC2xn − 2a2 + nC4xn−4a4 + nC6xn−6a6 + …….] 

Here, n = 6, x = √2, a = 1; 6C2 = 15, 6C4 = 15, 6C6 = 1 

(√2 + 1)6 + (√2 − 1)

= 2[(√2)6 + 15 . (√2)4 . 1 + 15(√2)2 . 1 + 1 . 1] 

= 2 [8 + 15 × 4 + 15 × 2 + 1]

= 198

Question 2: If (1 + ax)n = 1 + 8x + 24x2 + …., then the value of a and n is

Solution: 

As given (1 + ax)n = 1 + 8x + 24x2 + …. 

1 + (n / 1) ax + n (n − 1) / [1 . 2] a2x2 + ….

= 1 + 8x + 24x2 + …. 

⇒ na = 8, n (n − 1) / [1 . 2] a2 = 24

⇒ na (n − 1) a = 48 

8 (8 − a) = 48 

8 − a = 6

⇒ a = 2

⇒ n = 4

Question 3: The middle term in the expansion of (1 + 3x + 3x2 + x3)6 is

Solution: 

(1 + 3x + 3x2 + x3)6 = {(1 + x)3}

= (1 + x)18 

Hence the middle term of the expansion is the 10th term.

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