BINOMIAL EXPRESSION Before learning binomial expansion, let us recall what is a "binomial". A binomial is a polynomial with two terms. In other words, an algebraic expression consists of two dissimilar terms, connected by either a plus (+) or a minus (-) sign is called Binomial Expression. For example, a + b, x – y and 1 2 + 3 4 etc. are binomials. BINOMIAL THEOREM According to the binomial theorem, it is possible to expand any non-negative power of binomial (a + b) into a sum of the form, Where, n is a positive integer and a, b are real numbers, and 0 < r ≤ n. Each nCr is a positive integer known as a binomial coefficient. (a+b)n = nC0 a nb 0 + nC1 a n-1b 1 + nC2 a n-2 b 2 +........................+ nCn-1 a 1b n-1 + nCn a 0b n This formula helps to expand the binomial expressions such as (x + a)10, (2x + 5)3 , (x - (1/x))4 , and so on. The binomial theorem formula helps in the expansion of a binomial raised to a certain power. Let’s illustrate the above theorem with the help of an example as under: The first term's i.e. a, exponents start at “n” and go down (3,2,1,0) The second term's i.e. b, exponents start at “0” and go up (0,1,2,3)
• Coefficients are from Pascal's Triangle, or by calculation using n! r!(n−r)! NUMBER OF TERMS IN THE BINOMIAL EXPANSION The number of terms in the binomial expansion is determined by the power (exponent) of the binomial and is given by the (n + 1). In the binomial expansion of (a + b)n, where n is the power of the binomial, there will be (n + 1) number of terms. Let’s take an example to illustrate the above: Binomial Expansion (a + b)n N No. of Terms (a + b)2 = a2 + 2ab + b2 n = 2 (n+1) = 2+1 = 3 (a + b)3 = a3 + 3a2b + 3ab2 + b3 n = 3 (n+1) = 3+1 = 4 (a + b)4 = a4 + 4a3b + 6a2b 2 +4ab3 + b 4 n = 4 (n+1) = 4+1 = 5 Binomial Coefficients: Binomial coefficients, often denoted as nCr or C(n, r), where 'n' represents the total number of items in the set, ‘r' represents the number of items you want to choose and C stands for Combinations. For example, let's say you have a set of 5 different fruits: apple, banana, cherry, date, and elderberry. If you want to choose 2 fruits from this set, the binomial coefficient C(5, 2) tells you how many different combinations of 2 fruits you can make. To calculate C (5, 2), we use the formula nCr = n! r!(n−r)! where '!' denotes factorial. In simple terms, C (5, 2) means "choose 2 fruits from a set of 5 fruits." The calculation gives you the answer 10, which means there are 10 different combinations of 2 fruits you can make from the given set. Here are 10 specific combinations of 2 fruits that you can choose from a set of 5 fruits (apple, banana, cherry, date, and elderberry):
1. Apple + Banana 2. Apple + Cherry 3. Apple + Date 4. Apple + Elderberry 5. Banana + Cherry 6. Banana + Date 7. Banana + Elderberry 8. Cherry + Date 9. Cherry + Elderberry 10. Date + Elderberry As you can see, there are indeed 10 different combinations of 2 fruits that can be made from the given set. You can verify that each combination is unique and that there are no repetitions. Relationship between Binomial Expansion and Pascal Triangle The numbers in Pascal’s triangle form the coefficients in the binomial expansion. When we expand the expression (a + b)n using the binomial theorem, the coefficients of each term in the expansion can be found in the 'n+1'th row of Pascal's Triangle. Let's take an example to make it clearer. To expand this expression, we use the binomial theorem formula: Expand (a + b)n (a + b)n = C(4, 0) * a4 * b0 + C(4, 1) * a3 * b1 + C(4, 2) * a2 * b2 + C(4, 3) * a1 * b3 + C(4, 4) * a0 * b4 . (a+b)4 = a4 + 4a3b + 6a2b 2 + 4ab3 + b4 Now, let's look at the '4+1' = 5th row of Pascal's Triangle: 1 4 6 4 1. Comparing the coefficients in the expansion formula with the numbers in the 5th row of Pascal's Triangle, we see that they match: C(4, 0) = 1 C(4, 1) = 4 C(4, 2) = 6 C(4, 3) = 4 C(4, 4) = 1 So, the coefficients in the expansion of (a+b)4 match the numbers in the '4+1' = 5th row of Pascal's Triangle. This relationship between the coefficients and Pascal's Triangle
holds true for any power of (a + b). The 'n+1'th row of Pascal's Triangle corresponds to the coefficients in the expansion of (a + b)n . BINOMIAL EXPANSION VISA-VIS PASCAL TRIANGLE Ascertaining Specific Term of the Binomial Expansion To find a specific term, we rely on the concept of the (r+1)th term, where "r" represents the position or index of the term we are interested in. In the expansion of (a + b) n , the T(r+1)th term can be ascertained as under: +1 = × −×
Here, "n" signifies the exponent or power of the binomial expression, while "r" determines the position of the term we are interested in. It is essential to note that indexing conventionally commences at 0. Tr+1 refers to the term at position r+1 in the expansion. Each term in the expansion is denoted by "T" followed by the index or position of the term. So, in the above expression "T" refers to the general term of the binomial expansion. To clarify further, in the expansion of (a + b) n , the terms at different positions are as follows: T0 = The first term a 4 T1 = The second term 4a3b T2 = The third term 6a2b 2 T3 = The fourth term 4ab3 T4 = The fifth term b 4 For example, Find the 3rd term in the expansion of (a + b)4 , we need to consider the term at position 2, as the index convention starts from 0. We set r = 2. Applying the formula, we obtain: T3 = C(4, 2) * a(4-2) * b2 = 6a 2b 2 Alternatively, we can find r as under r+1= 3, therefore, r=2. PROPERTIES OF BINOMIAL COFFICIENT 1. nCn = nC0 = 1 The property states that the binomial coefficients nCn and nC0 are both equal to 1. This means that when choosing all objects (nCn) or choosing none (nC0) from a set will give value 1.
Example: Let's consider an example with n = 5. • nCn = 5C5 = 1. This means choosing all 5 objects from a set of 5, and there is only one way to do that: selecting the entire set. • nC0 = 5C0 = 1. This means choosing 0 objects from a set of 5, and there is only one way to do that: not selecting any object. 2. nC1 = nCn-1 = n This property states that the binomial coefficients n C1and n Cn-1 are both equal to the value of n itself. Example: Let's take an example with n = 4. • nC1 = 4C1 = 4. This means choosing 1 object from a set of 4, and there are four ways to do this: selecting any one of the four objects individually. • nC(n-1) = 4C(4-1) = 4C3 = 4. This means choosing 3 objects from a set of 4, leaving out one object. Again, there are four ways to do this: choosing any combination of three objects from the set. In both cases, the result is equal to n, which is 4 in this example. 3. nCr = nCn-r for r>0 In simple terms, this property tells us that choosing r objects from a set of n objects is equivalent to choosing (r-1) objects from the same set. Using this formula, let’s calculate: 5C3 and 5C2 • 5C3 = 5! / (3! * (5-3)!) = 5! / (3! * 2!) = (5 * 4 * 3!) / (3! * 2 * 1) = 10 So, there are 10 different ways to choose a committee of 3 students from a group of 5. • 5C2 = 5! / (2! * (5-2)!) = 5! / (2! * 3!) = (5 * 4 * 3!) / (2 * 1 * 3!) = 10 5C3 = 5C2
Series Summation (Binomial Coefficients) 1. C0+ C1 + C2 + C3 + C4 + .......Cn = 2n In this property, we sum up all the binomial coefficients from C0 to Cn, and the result is equal to 2 n and it means that the sum of all the binomial coefficients in the expansion of (a + b)^n is equal to 2 n . Example: Consider the expansion of (a + b)4 : C0 + C1 + C2 + C3 + C4 = 1 + 4 + 6 + 4 + 1 = 16 which is equal to 2 4 2. C0 + C2 + C4 + .... = C1 + C3 + C5 + ....... = 2n-1 It means that the sums of the even-indexed and odd-indexed binomial coefficients in the expansion of (a + b)^n are both equal to 2 n-1 . Example: Consider the expansion of (a + b)4 : let's calculate the sum of the even-indexed binomial coefficients: C0 + C2 + C4 = 1 + 6 + 1 = 8 Next, let's calculate the sum of the odd-indexed binomial coefficients: C1 + C3 = 4 + 4 = 8 As you can see, the sum of the even-indexed binomial coefficients (C0 + C2 + C4) and the sum of the odd-indexed binomial coefficients (C1 + C3) are both equal to 8, which is equal to 24-1 (2(n-1)) 3. C0 - C1 + C2 - C3 + C4 - C5 + .... = 0 This property relates to the alternating signs of the binomial coefficients in the expansion of (a + b)4 . Example: let's consider the expansion of (a + b)^3:
In this expansion, the binomial coefficients alternate in sign: positive for C0 and C2, and negative for C1 and C3. Let's calculate the sum of these alternating binomial coefficients: C0 - C1 + C2 - C3 = 1 - 3 + 3 - 1 = 0 The property C0 - C1 + C2 - C3 + C4 - C5 + ... = 0 suggests that the binomial coefficients with even indices (like C0, C2, C4, etc.) will have positive signs, while the binomial coefficients with odd indices (like C1, C3, C5, etc.) will have negative signs. This leads to the cancellation of these alternating terms, resulting in the sum being equal to 0. Note: The alternating signs occur when expanding the binomial (a - b)^n, where the signs of the terms alternate between positive and negative.