In the previous section, we completed a discussion on permutations and combinations. In this chapter, we will see Binomial theorem.
Some basics can be written in 4 steps:
1. In our earlier classes, we have seen the basics about polynomials (Details here and here).
• Now, a binomial is a polynomial having two terms.
♦ In some binomials the two terms will be added together.
♦ In some binomials, one term will be subtracted from the other term.
• Some examples of binomials are: (a+b), (a-b), (2x2-3), (4x+5) etc.,
2. In many scientific and engineering problems, we will want to find squares, cubes and higher powers of binomials. Let us see some examples:
(i) (4x+5)2 is the square of the binomial (4x+5)
♦ It can be easily calculated using the identity: (a+b)2 = a2+2ab+b2
• Here, the binomial is raised to the power 2
• Power is also known as index (plural indices)
(ii) (4x+5)3 is the cube of the binomial (4x+5)
♦ It can be easily calculated using the identity: (a+b)3 = a3+3a2b+3ab2+b3
• Here, the binomial is raised to the power 3
(iii) When the binomial (4x+5) is raised to the power 4, we write: (4x+5)4
♦ It can be calculated using the identity:
(a+b)4 = (a+b)(a+b)3 = (a+b)(a3+3a2b+3ab2+b3) = a4+4a3b+6a2b2+4ab3+b4
• Here, the binomial is raised to the power 4
(iv) When the binomial is raised to the power 5, we write: (4x+5)5
so on . . .
• From the 5th power on wards, the calculations become lengthy.
3. In general, when a binomial is raised to the power n, we write (a+b)n
♦ n can be an integer like -3, 2, -5 etc.,
♦ n can be a rational number like $-\frac{1}{3},~\frac{2}{5}$ etc.,
4. We will encounter situations where,
♦ n is very large like 12, 25, -14 etc.,
♦ n is rational numbers like $-\frac{1}{3},~\frac{2}{5}$ etc.,
◼ In such cases, the binomial theorem helps us to reduce calculation steps.
• In this chapter we will see how binomial theorem helps us when n is a +ve integer.
• In higher classes, we will see the cases where n is -ve integer or rational number.
Pascal's Triangle
Binomial theorem is based on Pascal's Triangle. It's details can be written in 3 steps:
1. Consider the basic identities:
♦ When n = 0, (a+b)n = (a+b)0 = 1 [Here (a+b) should not be equal to zero]
♦ When n = 1, (a+b)n = (a+b)1 = (a+b)
♦ When n = 2, (a+b)n = (a+b)2 = a2+2ab+b2
♦ When n = 3, (a+b)n = (a+b)3 = a3+3a2b+3ab2+b3
♦ When n = 4, (a+b)n = (a+b)4 = a4+4a3b+6a2b2+4ab3+b4
2. In the above expansions, we see three peculiarities:
(i) The total number of terms in the expansion, is one more than the index.
• For example, consider (a+b)3:
The index is 3 and the number of terms in the expansion is 4
(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the
second quantity ‘b’ increase by 1, in the successive terms.
• For example, consider (a+b)4 = a4+4a3b+6a2b2+4ab3+b4 :
♦ In the first term, power of a is 4 and that of b is 0
♦ In the second term,
✰ the power of a decreases by 1 and becomes 3
✰ the power of b increases by 1 and becomes 1
♦ In the third term,
✰ the power of a decreases by 1 and becomes 2
✰ the power of b increases by 1 and becomes 2
♦ In the fourth term,
✰ the power of a decreases by 1 and becomes 1
✰ the power of b increases by 1 and becomes 3
♦ In the fifth term,
✰ the power of a decreases by 1 and becomes 0
✰ the power of b increases by 1 and becomes 4
(iii) In each term of the expansion, the sum of the indices of a and b is the same and is equal to the index of (a+b).
• For example, consider (a+b)4 = a4+4a3b+6a2b2+4ab3+b4 :
♦ In the first term, the sum of the indices is (4+0) = 4
✰ Sum is same as the index of (a+b)4.
♦ In the second term, the sum of the indices is (3+1) = 4
✰ Sum is same as the index of (a+b)4.
♦ In the third term, the sum of the indices is (2+2) = 4
✰ Sum is same as the index of (a+b)4.
♦ In the fourth term, the sum of the indices is (1+3) = 4
✰ Sum is same as the index of (a+b)4.
♦ In the fifth term, the sum of the indices is (0+4) = 4
✰ Sum is same as the index of (a+b)4.
3. Now we will see an interesting relation between the coefficients of the expansions. It can be written in 4 steps:
(i) Fig.8.1 below shows the coefficients written in order.
• The first row shows the coefficients when (a+b) is raised to the power zero.
♦ We know that, any number raised to zero will give 1.
♦ So there is a one in the first row.
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| Fig.8.1 |
• The second row shows the coefficients when (a+b) is raised to the power 1.
♦ We know that, (a+b) raised to 1 will give the same (a+b).
✰ Coefficient of the first term is 1.
✰ Coefficient of the second term is 1.
♦ Thus there are two ones in the second row.
• The third row shows the coefficients when (a+b) is raised to the power 2.
♦ We know that, (a+b) raised to 2 will give a2+2ab+b2.
✰ Coefficient of the first term is 1.
✰ Coefficient of the second term is 2.
✰ Coefficient of the third term is 1.
♦ Thus in the third row, we have 1, 2, 1
• The fourth row shows the coefficients when (a+b) is raised to the power 3.
♦ We know that, (a+b) raised to 3 will give a3+3a2b+3ab2+b3.
✰ Coefficient of the first term is 1.
✰ Coefficient of the second term is 3.
✰ Coefficient of the third term is 3.
✰ Coefficient of the fourth term is 1.
♦ Thus in the fourth row, we have 1, 3, 3, 1
• The fifth row shows the coefficients when (a+b) is raised to the power 4.
♦ We know that, (a+b) raised to 4 will give a4+4a3b+6a2b2+4ab3+b4.
✰ Coefficient of the first term is 1.
✰ Coefficient of the second term is 4.
✰ Coefficient of the third term is 6.
✰ Coefficient of the fourth term is 4.
✰ Coefficient of the fifth term is 1.
♦ Thus in the fifth row, we have 1, 4, 6, 4, 1
(ii) We can easily write the coefficients up to the row where the index is 4.
♦ For writing the next row, we first will need to expand (a+b)5
♦ For writing the row below that, we will first need to expand (a+b)6.
♦ So on . . .
(iii) This is a lengthy process. So we must find an easier way.
◼ In fig.8.1, we see a pattern:
♦ Adding the two ones in index 1, will give the middle coefficient in index 2.
♦ This is indicated by the top most triple headed arrow in fig.8.2 below:
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| Fig.8.2 |
♦ Adding 1 and 2 of index 2 will give the first 3 in index 3
♦ Adding 2 and 1 of index 2 will give the second 3 in index 3
♦ This is indicated by the two triple headed arrows in the third row (row for index 2).
• We can continue like this and find the coefficients for higher indices.
(iv) The structure in fig.8.2 looks like a triangle.
♦ At the apex of the triangle, there is a ‘1’
♦ Along the left sloping side side of the triangle, all numbers are ‘1’
♦ Along the right sloping side of the triangle, all numbers are ‘1’
• This array of numbers is known as Pascal’s triangle. This name is given in honor of the French Mathematician Blaise Pascal.
Let us see an example:
Expand (2x+3y)5 using Pascal’s triangle
Solution:
1. Based on the peculiarities that we wrote in step (2) above,
♦ In the first term, the first quantity (2x) will have a power of 5
♦ In the first term, the second quantity (3y) will have a power of 0
• Using the Pascal's triangle in fig.8.2, we get:
♦ In the first term, the coefficient will be '1'
• Combining these three information, we get:
First term = 1 × (2x)5 × (3y)0 = 32x5
2. Based on the peculiarities that we wrote in step (2) above,
♦ In the second term, the first quantity (2x) will have a power of 4
♦ In the second term, the second quantity (3y) will have a power of 1
• Using the Pascal's triangle in fig.8.2, we get:
♦ In the second term, the coefficient will be '5'
• Combining these three information, we get:
Second term = 5 × (2x)4 × (3y)1 = 240x4y
3. Based on the peculiarities that we wrote in step (2) above,
♦ In the third term, the first quantity (2x) will have a power of 3
♦ In the third term, the second quantity (3y) will have a power of 2
• Using the Pascal's triangle in fig.8.2, we get:
♦ In the third term, the coefficient will be '10'
• Combining these three information, we get:
Third term = 10 × (2x)3 × (3y)2 = 720x3y2
4. In this way, we can write all the terms. We get:
(2x+3y)5 = 32x5 + 240x4y + 720x3y2 + 1080x2y3+ 810xy4+ 243y5
• We have successfully written the expansion of (2x+3y)5.
• What if we want the expansion of (2x+3y)12 ?
♦ We will have to write all the rows of the Pascal's triangle up to index 12.
♦ This is a lengthy process.
• So we must develop a rule that will help us to write any row in the Pascal's triangle, with out writing the rows above it.
• We will see such a rule in the next section.
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