Chapter 7.1: Introduction to Binomial Theorem
- Binomial Expression: An expression of the form (a + b)ⁿ, where n is a non-negative integer.
- Binomial Theorem: A formula that expands a binomial expression raised to a power.
Chapter 7.2: General Term in the Expansion of (a + b)ⁿ
- General Term: The rth term in the expansion of (a + b)ⁿ is given by:
- Tᵣ = nCr * a^(n-r) * b^r
Chapter 7.3: Pascal’s Triangle
- Pascal’s Triangle:
- Coefficients in Binomial Expansion: The coefficients in the expansion of (a + b)ⁿ can be found using Pascal’s Triangle.
Key Concepts:
- Binomial expansion
- General term in binomial expansion
- Pascal’s Triangle
- Applications of the Binomial Theorem (e.g., finding specific terms in an expansion, approximating values)
Exercise 7.1
Expand each of the expressions in Exercises 1 to 5.
1. (1–2x) 5
Ans :
o expand the expression (1 – 2x)^5, we can use the Binomial Theorem, which states:
(a + b)^n = nC0 * a^n + nC1 * a^(n-1) * b + nC2 * a^(n-2) * b^2 + … + nCn * b^n
where nCk is the binomial coefficient, given by:
nCk = n! / (k!(n – k)!)
In this case, a = 1, b = -2x, and n = 5. Substituting these values into the Binomial Theorem formula, we get:
(1 – 2x)^5 = 5C0 * 1^5 * (-2x)^0 + 5C1 * 1^4 * (-2x)^1 + 5C2 * 1^3 * (-2x)^2 + 5C3 * 1^2 * (-2x)^3 + 5C4 * 1^1 * (-2x)^4 + 5C5 * 1^0 * (-2x)^5
Now, let’s calculate each term:
- 5C0 * 1^5 * (-2x)^0 = 1
- 5C1 * 1^4 * (-2x)^1 = -10x
- 5C2 * 1^3 * (-2x)^2 = 40x^2
- 5C3 * 1^2 * (-2x)^3 = -80x^3
- 5C4 * 1^1 * (-2x)^4 = 80x^4
- 5C5 * 1^0 * (-2x)^5 = -32x^5
Finally, combining all the terms:
(1 – 2x)^5 = 1 – 10x + 40x^2 – 80x^3 + 80x^4 – 32x^5
Therefore, the expansion of (1 – 2x)^5 is:
1 – 10x + 40x^2 – 80x^3 + 80x^4 – 32x^5
2.
Ans :
3. (2x – 3)6
Ans :
4.
Ans :
5.
Ans :
Using binomial theorem, evaluate each of the following
6. (96)3
Ans :
7. (102)5
Ans :
In this case, a = 100, b = 2, and n = 5. Substituting these values into the binomial theorem formula, we get:
(100 + 2)^5 = 5C0 * 100^5 * 2^0 + 5C1 * 100^4 * 2^1 + 5C2 * 100^3 * 2^2 + 5C3 * 100^2 * 2^3 + 5C4 * 100^1 * 2^4 + 5C5 * 100^0 * 2^5
Now, let’s calculate each term:
- 5C0 * 100^5 * 2^0 = 1 * 10000000000 * 1 = 10000000000
- 5C1 * 100^4 * 2^1 = 5 * 100000000 * 2 = 1000000000
- 5C2 * 100^3 * 2^2 = 10 * 1000000 * 4 = 40000000
- 5C3 * 100^2 * 2^3 = 10 * 10000 * 8 = 800000
- 5C4 * 100^1 * 2^4 = 5 * 100 * 16 = 80000
- 5C5 * 100^0 * 2^5 = 1 * 1 * 32 = 32
Finally, combining all the terms:
(102)^5 = 10000000000 + 1000000000 + 40000000 + 800000 + 80000 + 32 = 11040808032
Therefore, (102)^5 is equal to 11,040,808,032.
8. (101)4
Ans :
In this case, a = 100, b = 1, and n = 4. Substituting these values into the binomial theorem formula, we get:
(100 + 1)^4 = 4C0 * 100^4 * 1^0 + 4C1 * 100^3 * 1^1 + 4C2 * 100^2 * 1^2 + 4C3 * 100^1 * 1^3 + 4C4 * 100^0 * 1^4
Now, let’s calculate each term:
- 4C0 * 100^4 * 1^0 = 1 * 100000000 * 1 = 100000000
- 4C1 * 100^3 * 1^1 = 4 * 1000000 * 1 = 4000000
- 4C2 * 100^2 * 1^2 = 6 * 10000 * 1 = 60000
- 4C3 * 100^1 * 1^3 = 4 * 100 * 1 = 400
- 4C4 * 100^0 * 1^4 = 1 * 1 * 1 = 1
Finally, combining all the terms:
(101)^4 = 100000000 + 4000000 + 60000 + 400 + 1 = 104060401
Therefore, (101)^4 is equal to 104,060,401.
9. (99)5
Ans :
In this case, a = 100, b = -1, and n = 5. Substituting these values into the binomial theorem formula, we get:
(100 – 1)^5 = 5C0 * 100^5 * (-1)^0 + 5C1 * 100^4 * (-1)^1 + 5C2 * 100^3 * (-1)^2 + 5C3 * 100^2 * (-1)^3 + 5C4 * 100^1 * (-1)^4 + 5C5 * 100^0 * (-1)^5
Now, let’s calculate each term:
- 5C0 * 100^5 * (-1)^0 = 1 * 10000000000 * 1 = 10000000000
- 5C1 * 100^4 * (-1)^1 = 5 * 100000000 * (-1) = -500000000
- 5C2 * 100^3 * (-1)^2 = 10 * 1000000 * 1 = 10000000
- 5C3 * 100^2 * (-1)^3 = 10 * 10000 * (-1) = -100000
- 5C4 * 100^1 * (-1)^4 = 5 * 100 * 1 = 500
- 5C5 * 100^0 * (-1)^5 = 1 * 1 * (-1) = -1
Finally, combining all the terms:
(99)^5 = 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1 = 9509900501
Therefore, (99)^5 is equal to 9,509,900,501.
10. Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
Ans :
11.
Ans :
12.
Ans :
13. Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
Ans :
14.
Ans :
