IIT JEE Main Maths -Unit 5- Simple applications- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 5- Simple applications – Study Notes – New syllabus
IIT JEE Main Maths -Unit 5- Simple applications – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Binomial Theorem for Any Rational Index (Generalized Binomial Expansion)
- Binomial Theorem — Approximations and Applications (Including Negative and Fractional Indices)
- Binomial Theorem — Properties and Summation of Binomial Coefficients
Binomial Theorem for Any Rational Index (Generalized Binomial Expansion)
The standard binomial theorem \( (x + a)^n = \sum_{r=0}^{n} \binom{n}{r}x^{n – r}a^r \) applies only when \( n \) is a positive integer.
However, the theorem can be extended to cases where the index \( n \) is any rational number (positive, negative, or fractional).
Generalized Binomial Theorem
For any rational number \( n \) and \( |x| < 1 \):
$ (1 + x)^n = 1 + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots $
where the binomial coefficients are defined as:
$ \binom{n}{r} = \dfrac{n(n – 1)(n – 2)\dots(n – r + 1)}{r!} $
This series is infinite when \( n \) is not a positive integer.
Expansion Examples
$ (1 + x)^{1/2} = 1 + \dfrac{1}{2}x – \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 – \dots $
$ (1 – x)^{-1} = 1 + x + x^2 + x^3 + x^4 + \dots $
$ (1 – x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots $
Validity of the Expansion
The series converges only for \( |x| < 1 \). For \( |x| \ge 1 \), the infinite series diverges (does not have a finite sum).
General Term in the Expansion
The general term (r-th term, starting from \( r = 0 \)) is:
$ T_{r+1} = \binom{n}{r}x^r = \dfrac{n(n – 1)(n – 2)\dots(n – r + 1)}{r!}x^r $
Important Special Cases
- Case 1: \( (1 + x)^{-1} = 1 – x + x^2 – x^3 + \dots \)
- Case 2: \( (1 – x)^{-1} = 1 + x + x^2 + x^3 + \dots \)
- Case 3: \( (1 + x)^{-2} = 1 – 2x + 3x^2 – 4x^3 + \dots \)
Example
Expand \( (1 + x)^{1/2} \) up to the term containing \( x^3 \).
▶️ Answer / Explanation
We know that:
\( (1 + x)^{1/2} = 1 + \dfrac{1}{2}x + \dfrac{(1/2)(-1/2)}{2!}x^2 + \dfrac{(1/2)(-1/2)(-3/2)}{3!}x^3 + \dots \)
\( = 1 + \dfrac{1}{2}x – \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + \dots \)
Answer: \( (1 + x)^{1/2} = 1 + \dfrac{1}{2}x – \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + \dots \)
Example
Find the coefficient of \( x^3 \) in the expansion of \( (1 – 2x)^{-3/2} \).
▶️ Answer / Explanation
General term: \( T_{r+1} = \binom{-3/2}{r}(-2x)^r \)
Coefficient of \( x^3 \): \( r = 3 \)
\( \binom{-3/2}{3} = \dfrac{(-3/2)(-5/2)(-7/2)}{3!} = -\dfrac{35}{16} \)
Hence, coefficient = \( -\dfrac{35}{16}(-2)^3 = -\dfrac{35}{16}(-8) = \dfrac{280}{16} = 17.5 \)
Answer: Coefficient of \( x^3 = 17.5 \)
Example
Use the binomial expansion to approximate \( (1.02)^5 \) correct up to 3 decimal places.
▶️ Answer / Explanation
Let \( (1 + x)^5 \) with \( x = 0.02 \).
\( (1 + 0.02)^5 = 1 + 5(0.02) + \dfrac{5 \times 4}{2!}(0.02)^2 + \dfrac{5 \times 4 \times 3}{3!}(0.02)^3 \)
= \( 1 + 0.1 + 0.008 + 0.0004 = 1.1084 \)
Answer: \( (1.02)^5 \approx 1.108 \)
Binomial Theorem for Negative and Fractional Indices
When the index \( n \) is negative or fractional, the expansion of \( (1 + x)^n \) is an infinite series that converges only for \( |x| < 1 \).
$ (1 + x)^n = 1 + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots $
where the generalized binomial coefficients are:
$ \binom{n}{r} = \dfrac{n(n – 1)(n – 2)\dots(n – r + 1)}{r!}, \quad r = 0, 1, 2, \dots $
Common Expansions (to Remember for JEE)
| Expression | Binomial Expansion |
|---|---|
| \( (1 + x)^{-1} \) | \( 1 – x + x^2 – x^3 + x^4 – \dots \) |
| \( (1 – x)^{-1} \) | \( 1 + x + x^2 + x^3 + \dots \) |
| \( (1 + x)^{-2} \) | \( 1 – 2x + 3x^2 – 4x^3 + 5x^4 – \dots \) |
| \( (1 + x)^{1/2} \) | \( 1 + \dfrac{1}{2}x – \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 – \dots \) |
General Term in the Expansion
The general term (\( r + 1 \)th term) is given by:
$ T_{r+1} = \binom{n}{r}x^r = \dfrac{n(n – 1)(n – 2)\dots(n – r + 1)}{r!}x^r $
Convergence Condition
The binomial series for negative or fractional index is valid only when:
$ |x| < 1 $
This ensures that the infinite series converges.
Approximations for Small Values of \( x \)
When \( |x| \ll 1 \), we can approximate \( (1 + x)^n \) by retaining only a few terms:
$ (1 + x)^n \approx 1 + nx + \dfrac{n(n – 1)}{2}x^2 $
This is extremely useful for approximations and error estimation in JEE.
Applications of Binomial Theorem (JEE-Oriented)
- Approximating values like \( (1.02)^5, (0.98)^3, \dfrac{1}{(1.03)^2} \), etc.
- Solving limit problems where \( x \) tends to 0.
- Error estimation in small measurements or physical quantities.
- Simplification of algebraic expressions using first few terms of expansion.
Example
Expand \( (1 – x)^{-3} \) up to the term containing \( x^3 \).
▶️ Answer / Explanation
\( (1 – x)^{-3} = 1 + (-3)(-x) + \dfrac{(-3)(-4)}{2!}(-x)^2 + \dfrac{(-3)(-4)(-5)}{3!}(-x)^3 + \dots \)
= \( 1 + 3x + 6x^2 + 10x^3 + \dots \)
Answer: \( (1 – x)^{-3} = 1 + 3x + 6x^2 + 10x^3 + \dots \)
Example
Approximate \( \dfrac{1}{(1.02)^3} \) up to three decimal places using binomial expansion.
▶️ Answer / Explanation
We can write \( \dfrac{1}{(1.02)^3} = (1 + 0.02)^{-3} \).
\( (1 + x)^{-3} = 1 – 3x + 6x^2 – 10x^3 + \dots \)
Substitute \( x = 0.02 \): \( = 1 – 3(0.02) + 6(0.02)^2 – 10(0.02)^3 = 1 – 0.06 + 0.0024 – 0.00008 = 0.94232 \)
Answer: \( \dfrac{1}{(1.02)^3} \approx 0.942 \)
Example
Find the first four terms in the expansion of \( (1 – 2x)^{-1/2} \).
▶️ Answer / Explanation
\( (1 – 2x)^{-1/2} = 1 + \dfrac{-1}{2}(-2x) + \dfrac{(-1/2)(-3/2)}{2!}(-2x)^2 + \dfrac{(-1/2)(-3/2)(-5/2)}{3!}(-2x)^3 + \dots \)
= \( 1 + x + \dfrac{3}{2}x^2 + \dfrac{5}{2}x^3 + \dots \)
Answer: \( (1 – 2x)^{-1/2} = 1 + x + \dfrac{3}{2}x^2 + \dfrac{5}{2}x^3 + \dots \)
Binomial Theorem — Properties and Summation of Binomial Coefficients (JEE Shortcuts and Patterns)
Definition of Binomial Coefficients
The binomial coefficients are the constants \( \displaystyle \binom{n}{r} \) that appear in the expansion of \( (x + a)^n \):
$ (x + a)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n – r} a^r $
where
$ \binom{n}{r} = \dfrac{n!}{r!(n – r)!}, \quad r = 0, 1, 2, \dots, n $
Basic Properties of Binomial Coefficients
| Property | Formula |
|---|---|
| (i) Symmetry Property | \( \binom{n}{r} = \binom{n}{n – r} \) |
| (ii) Pascal’s Identity | \( \binom{n}{r} + \binom{n}{r – 1} = \binom{n + 1}{r} \) |
| (iii) First and Last Coefficient | \( \binom{n}{0} = \binom{n}{n} = 1 \) |
| (iv) Greatest Coefficient | If \( n \) is even → greatest = \( \binom{n}{n/2} \); if \( n \) is odd → two equal greatest coefficients \( \binom{n}{(n-1)/2} = \binom{n}{(n+1)/2} \) |
| (v) Sum Property | \( \sum_{r=0}^{n} \binom{n}{r} = 2^n \) |
Alternate Sum Property
$ \sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0 $
Proof: Expand \( (1 – 1)^n = 0 \Rightarrow \sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0 \).
Even and Odd Term Sums
$ \text{Sum of even terms} = \text{Sum of odd terms} = 2^{n-1} $
Proof:
$ (1 + 1)^n = 2^n \quad \text{and} \quad (1 – 1)^n = 0 $
$ \Rightarrow \text{Even + Odd} = 2^n, \quad \text{Even − Odd} = 0 $
$ \Rightarrow \text{Even} = \text{Odd} = 2^{n-1} $
General Summation Results
- \( \displaystyle \sum_{r=0}^{n} r\binom{n}{r} = n \times 2^{n – 1} \)
- \( \displaystyle \sum_{r=0}^{n} r^2\binom{n}{r} = n(n + 1) \times 2^{n – 2} \)
- \( \displaystyle \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \)
Binomial Theorem Applications in Coefficient Patterns
For expansions like \( (1 + x)^n + (1 – x)^n \):
$ (1 + x)^n + (1 – x)^n = 2\left[\binom{n}{0} + \binom{n}{2}x^2 + \binom{n}{4}x^4 + \dots\right] $
→ Contains only even powers of x.
For \( (1 + x)^n – (1 – x)^n \):
$ (1 + x)^n – (1 – x)^n = 2\left[\binom{n}{1}x + \binom{n}{3}x^3 + \dots\right] $
→ Contains only odd powers of x.
Example
Find the value of \( \displaystyle \sum_{r=0}^{5} \binom{5}{r} \).
▶️ Answer / Explanation
\( \sum_{r=0}^{5} \binom{5}{r} = 2^5 = 32 \)
Answer: 32
Example
Find \( \displaystyle \sum_{r=0}^{8} r\binom{8}{r} \).
▶️ Answer / Explanation
Using formula \( \sum r\binom{n}{r} = n2^{n-1} \):
\( = 8 \times 2^{7} = 8 \times 128 = 1024 \)
Answer: 1024
Example
Find \( \displaystyle \sum_{r=0}^{10} (-1)^r \binom{10}{r} \).
▶️ Answer / Explanation
We know that:
\( (1 – 1)^{10} = 0 \Rightarrow \sum_{r=0}^{10} (-1)^r \binom{10}{r} = 0 \)
Answer: 0
Example
Find \( \displaystyle \sum_{r=0}^{n} \binom{n}{r}^2 \).
▶️ Answer / Explanation
We know the identity:
\( \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \)
Answer: \( \displaystyle \binom{2n}{n} \)