Binomial Theorem Class 11 Notes AP Inter 1st Year Maths Chapter 7

Students can go through AP Inter 1st Year Maths Notes 7th Lesson Binomial Theorem will help students in revising the entire concepts quickly.

Binomial Theorem Class 11 Notes AP Inter 1st Year Maths 7th Lesson

→ The expansion of a binomial for any positive integer n is given by the Binomial Theorem, which is

→ The general term of (a + b)ⁿ is T_r+1 = ⁿC_r a^n-r b^r, (0 ≤ r ≤ n)

→ The middle term in the expansion (a + b)ⁿ is
If n is even, then the number of terms in the expansion is n + 1. Since n is even so n + 1 is odd. Therefore, the middle term is \(\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }}\) term.
If n is odd, then n + 1 is even, so there will be two middle terms in the expansion, namely \(\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}\) term and \(\left(\frac{\mathrm{n}+1}{2}+1\right)^{\text {th }}\) term.

→ The coefficients of the expansions are arranged in an array. This array is called Pascal’s triangle.

→ The coefficients in the binomial expansion are ⁿC₀, ⁿC₁, ⁿC₂, ….., ⁿC_r, ……., ⁿC_n. These coefficients are called the binomial coefficients (corresponding to n).

→ When n is fixed, these coefficients are usually denoted by C₀, C₁, C₂,…, C_r,…, C_n respectively.

Example Problems

Question 1.
Expand \(\left(x^2+\frac{3}{x}\right)^4\), x ≠ 0
Solution:
By using the binomial theorem, we have

Question 2.
Compute (98)⁵.
Solution:
We express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use the Binomial theorem.
Write 98 = 100 – 2
Therefore, (98)⁵ = (100 – 2)⁵
= ⁵C₀ (100)⁵ – ⁵C₁ (100)⁴ . 2 + ⁵C₂ (100)³ 2² – ⁵C₃ (100)² (2)³ + ⁵C₄ (100) (2)⁴ – ⁵C₅ (2)⁵
= 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 × 10000 × 8 + 5 × 100 × 16 – 32
= 10040008000 – 1000800032
= 9039207968

Question 3.
Which is larger (1.01)^1000000 or 10,000?
Solution:
Splitting 1.01 and using the binomial theorem to write the first few terms, we have
(1.01)^1000000 = (1 + 0.01)^1000000
= ^1000000C₀ + ^1000000C₁ (0.01) + Other positive terms
= 1 + 1000000 × 0.01 + other positive terms
= 1 + 10000 + other positive terms > 10000
Hence (1.01)^1000000 > 10000

Question 4.
Using the binomial theorem, prove that 6ⁿ – 5n always leaves a remainder of 1 when divided by 25.
Solution:
For two numbers a and b is we can find numbers q and r such that a = bq + r, then we say that b divides a with q as the quotient and r as the remainder.
Thus, to show that 6ⁿ – 5n leaves a remainder of 1 when divided by 25, we prove that 6ⁿ – 5n = 25k + 1, where k is some natural number. We have

This shows that when divided by 25, 6ⁿ – 5n leaves a remainder of 1.

Multiple Choice Questions

Question 1.
The 3rd term of \(\left(3 x-\frac{y^3}{6}\right)^4\) is

Answer:
(1) \({ }^4 C_2(3 x)^2\left(\frac{-y^3}{6}\right)^2\)

Question 2.
C₀ + 3.C₁ + 3².C₂ + …. + 3ⁿ.C_n = ________
(1) 2ⁿ
(2) 3ⁿ
(3) 4ⁿ
(4) 5ⁿ
Answer:
(3) 4ⁿ

Question 3.
The middle term in the expansion of (4 + 2x)⁶ is
(1) 11240x²
(2) 10240x³
(3) 12240x⁴
(4) 10340x⁴
Answer:
(2) 10240x³

Question 4.
The coefficient of x²y² in the expansion of (x + 1)² (y + 1)³ is
(1) 3
(2) 5
(3) 2
(4) 10
Answer:
(1) 3

Question 5.
In the expansion of (1 + x)²⁰, if the coefficients of r^th and (r + 4)^th terms are equal, then r is
(1) 7
(2) 8
(3) 9
(4) 10
Answer:
(3) 9

Question 6.
The remainder when 8⁴⁸ is divided by 63 is
(1) 4
(2) 2
(3) 1
(4) 7
Answer:
(3) 1

Question 7.
The expansion of (x + y)¹⁰⁰⁰ is

Answer:
(2) \(\sum_{r=0}^{1000}{ }^{1000} C_r x^{1000-r} y^r\)

Question 8.
The coefficients of the first and the last terms of the expansion (x + y)ⁿ are
(1) 2
(2) 3
(3) n
(4) 1
Answer:
(4) 1

Question 9.
The expansion of (xy + 2)² is
(1) x² + y² + 4
(2) x²y² + 4xy + 4
(3) xy² + 4 + 2xy
(4) x²y² + 2xy + 4
Answer:
(2) x²y² + 4xy + 4

Question 10.
The term ⁹¹C₂ x⁸⁹ belongs to
(1) x⁸⁹
(2) (x – 2)⁹⁰
(3) (x – 1)⁹¹
(4) (x + 1)⁹⁰
Answer:
(3) (x – 1)⁹¹

Question 11.
The coefficient of x⁸y¹⁰ in the expansion of (x + y)¹⁸ is
(1) ¹⁸C₈
(2) ¹⁸P₁₀
(3) 2¹⁸
(4) None of these
Answer:
(1) ¹⁸C₈