Chapter 8.1 - Binomial Theorem for Any Positive Integer

In the previous section, we saw the basic details about Pascal's triangle. We saw the necessity for developing a rule to find the various numbers in the Pascal's triangle. In this section, we will see such a rule.

The basics about the rule can be written in 6 steps:
1. We have seen the concept of combination in the previous chapter.
We saw the formula: ${}^n {\rm{C}}_r~=~\frac{n!}{r!(n-r)!}$
2. Let us check whether this formula is applicable to the Pascal's triangle. The triangle is shown below for easy reference:

Fig.8.2

The check can be done in 3 steps:
(i) Consider any coefficient in the Pascal's triangle, say the fourth coefficient for index 5
    ♦ The fourth coefficient of index 5 is 10
    ♦ ⁵C₃ is also 10
(ii) Consider any other coefficient in the Pascal's triangle, say the third coefficient for index 7
    ♦ The third coefficient of index 7 is 21
    ♦ ⁷C₂ is also 21
(iii) Let us check one more coefficient in the Pascal's triangle, say the fifth coefficient for index 8
    ♦ The fifth coefficient of index 8 is 70
    ♦ ⁸C₄ is also 70
3. This gives us an idea to pick out any coefficient in the Pascal’s triangle:
The x^th coefficient for any index n will be ⁿC_x-1
• For example, let n be 6 and x be 5
   ♦ Then the 5^th coefficient of index 6 will be be ⁶C_5-1 = ⁶C₄ = 15
   ♦ From the Pascal's triangle, we see that, this is true.
4. Based on this information, the Pascal’s triangle can be modified as shown in fig.8.3 below:

Fig.8.3

• In the modified triangle, we see two points:
(i) In any row, the subscript on the right side of ‘C’ progressively increases in the order: 0, 1, 2, 3, . . . n
   ♦ Where n is the index of that row.
(ii) In any row, the superscript on the left side of ‘C’ does not change.
   ♦ It is equal to n, the index of the row.
5. The above two points give us an idea to quickly write any row of the Pascal’s triangle.
• For example, based on the two points, the row for index 12 will be:
¹²C₀, ¹²C₁, ¹²C₂, ¹²C₃, ¹²C₄, ¹²C₅, ¹²C₆, ¹²C₇, ¹²C₈, ¹²C₉, ¹²C₁₀, ¹²C₁₁, ¹²C₁₂
6. If we can write any row of the Pascal’s triangle quickly, we will be able to write the expansion corresponding to that index also quickly.
• Let us see an example:
Expand (a+b)⁷
Solution:
1. The coefficients will be:
⁷C₀, ⁷C₁, ⁷C₂, ⁷C₃, ⁷C₄, ⁷C₅, ⁷C₆, and ⁷C₇
2. So the expansion will be:
(a+b)⁷ =
⁷C₀ a⁷ + ⁷C₁ a⁶b¹ + ⁷C₂ a⁵b² + ⁷C₃ a⁴b³ + ⁷C₄ a³b⁴ + ⁷C₅ a²b⁵ + ⁷C₆ a¹b⁶  + ⁷C₇ b⁷
• Remember that, the indices of a and b are calculated using the three peculiarities that we wrote in the previous section. They are given below for easy reference:

(i) The total number of terms in the expansion, is one more than the index. (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. (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).

• Thus we can write the complete expansion.

---

Now we can write the general form. It can be written in 3 steps:
(i) All the coefficients are of the form ⁿC_r
   ♦ For any particular problem, ‘n’ will be a constant.
         ✰ It will be equal to the index in the question.
   ♦ r will increase from 0 to n
(ii) The indices of a and b are calculated using the three peculiarities that we wrote in the previous section (also shown in the box above).
(iii) Based on this, the general form will be:
$(a+b)^n~=~{}^n {\rm{C}}_0 a^n~+~{}^n {\rm{C}}_1 a^{n-1} b~+~{}^n {\rm{C}}_2 a^{n-2} b^2~+~{}^n {\rm{C}}_3 a^{n-3} b^3~+~.~.~.~+~{}^n {\rm{C}}_{n-1} a b^{n-1}~+~{}^n {\rm{C}}_n b^n$

---

• Let us use this general form to expand (2x+5)⁶
We get:

(2x+5)⁶
$\begin{array}{ll} {}={}&{}^6 {\rm{C}}_0 (2x)^6\,5^0 &{}+{}& {}^6 {\rm{C}}_1 (2x)^5\,5^1 &{}+{}& {}^6 {\rm{C}}_2 (2x)^4\,5^2 &{}+{}& {}^6 {\rm{C}}_3 (2x)^3\,5^3 &{}+{}& {}^6 {\rm{C}}_4 (2x)^2\,5^4 &{}+{}& {}^6 {\rm{C}}_5 (2x)^1\,5^5 &{}+{}& {}^6 {\rm{C}}_6 (2x)^0\,5^6 \\ {}={}&{}^6 {\rm{C}}_0 2^6\,x^6\,5^0 &{}+{}& {}^6 {\rm{C}}_1 2^5\,x^5\,5^1 &{}+{}& {}^6 {\rm{C}}_2 2^4\,x^4\,5^2 &{}+{}& {}^6 {\rm{C}}_3 2^3\,x^3\,5^3 &{}+{}& {}^6 {\rm{C}}_4 2^2\,x^2\,5^4 &{}+{}& {}^6 {\rm{C}}_5 2^1\,x^1\,5^5 &{}+{}& {}^6 {\rm{C}}_6 2^0\,x^0\,5^6 \\ {}={}&{}^6 {\rm{C}}_0  × 64 × x^6 × 5^0 &{}+{}& {}^6 {\rm{C}}_1  × 32 × x^5 × 5^1 &{}+{}& {}^6 {\rm{C}}_2  × 16 × x^4\,5^2 &{}+{}& {}^6 {\rm{C}}_3  × 8 × x^3 × 5^3 &{}+{}& {}^6 {\rm{C}}_4  × 4 × x^2 × 5^4 &{}+{}& {}^6 {\rm{C}}_5  × 2 × x^1\,5^5 &{}+{}& {}^6 {\rm{C}}_6  × 1 × x^0\,5^6 \\ {}={}&1  × 64 × x^6 × 5^0 &{}+{}& 6  × 32 × x^5 × 5^1 &{}+{}& 15  × 16 × x^4 × 5^2 &{}+{}& 20  × 8 × x^3 × 5^3 &{}+{}& 15  × 4 × x^2 × 5^4 &{}+{}& 6  × 2 × x^1 × 5^5 &{}+{}& 1  × 1 × x^0 × 5^6 \\ {}={}&1  × 64 × x^6 × 1 &{}+{}& 6  × 32 × x^5 × 5 &{}+{}& 15  × 16 × x^4 × 25 &{}+{}& 20  × 8 × x^3 × 125 &{}+{}& 15  × 4 × x^2 × 625 &{}+{}& 6  × 2 × x^1 × 3125 &{}+{}& 1  × 1 × x^0 × 15625 \\ {}={}&64 x^6 &{}+{}& 960 x^5 &{}+{}& 6000 x^4 &{}+{}& 20000 x^3 &{}+{}& 37500 x^2 &{}+{}& 37500 x &{}+{}& 15625 \\ \end{array}$

---

We have seen the general form. The RHS of that general form is a bit lengthy. It can be easily shortened as explained by the 4 steps below:

1. We see that, the RHS is a sum of a finite number of terms. So it is a summation.
• We can use the symbol '∑' (Greek upper case letter sigma) to denote summation. Thus the general form becomes:
$(a+b)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;a^{n-k}\;b^k}$
2. We know that, for any particular problem, n will be a constant. It is the index.
3. k is not a constant. It takes different values in different terms.
• 'k=0' at the bottom of '∑' indicates that, the value of k starts from zero. In other words, the value of k in the first term is zero.
• 'k=n' at the top of '∑' indicates that, the value of k ends at n. In other words, the value of k in the last term is n.
• So the values of k are: 0, 1, 2, 3, . . . n
• It is known as the summation from k = 0 to k = n.
4. The index of a is (n-k). The index of b is k.
This shows that, the sum of the indices of a and b will be n in every term.

---

Let us see some special cases:
Case 1:
In this case, we investigate the expansion when the second quantity is -ve. This can be written in 3 steps:
1. Let us put a = x and b = -y. Then we get:
(a+b)ⁿ = (x-y)ⁿ = [x + (-y)]ⁿ
2. Expanding this using the general form, we get:
$[x+(-y)]^n$
$\begin{array}{ll} {}={}&{}^n {\rm{C}}_0 \,x^n\,(-y)^0 &{}+{}& {}^n {\rm{C}}_1\, x^{n-1}\,(-y)^1 &{}+{}& {}^n {\rm{C}}_2 \,x^{n-2}\,(-y)^2 &{}+{}& {}^n {\rm{C}}_3\, x^{n-3}\,(-y)^3 &{}+{}& .~.~.~+~{}^n {\rm{C}}_n\, x^0\,(-y)^n&{}& {} &{} \\ {}={}&{}^n {\rm{C}}_0 \,x^n\,(-1)^0\,y^0 &{}+{}& {}^n {\rm{C}}_1\, x^{n-1}\,(-1)^1\,y^1 &{}+{}& {}^n {\rm{C}}_2 \,x^{n-2}\,(-1)^2\,y^2 &{}+{}& {}^n {\rm{C}}_3\, x^{n-3}\,(-1)^3\,y^3 &{}+{}& .~.~.~+~{}^n {\rm{C}}_n\, x^0\,(-1)^n\,y^n&{}& {} &{} \\ \end{array}$
$~{}={}~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k \, x^{n-k}\;y^k}$
3. Based on the above result, we can write:
$(a-b)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k \, a^{n-k}\;b^k}$

◼ Let us expand (x-2y)⁵ using this result. We get:
$(x-2y)^5~=~[x+(-2y)]^5~=~\sum\limits_{k\,=\,0}^{k\,=\,5}{{}^5 {\rm{C}}_k\;(-1)^k \, x^{6-k}\;(-2y)^k}$
• Once we write the sigma notation on the RHS, we will be able to write all the terms easily. We get:
$(x-2y)^5$
$\begin{array}{ll} {}={}&{}^5 {\rm{C}}_0 \,x^5\,(-1)^0\,(2y)^0 &{}+{}& {}^5 {\rm{C}}_1\, x^{5-1}\,(-1)^1\,(2y)^1 &{}+{}& {}^5 {\rm{C}}_2 \,x^{5-2}\,(-1)^2\,(2y)^2 &{}+{}& {}^5 {\rm{C}}_3\, x^{5-3}\,(-1)^3\,(2y)^3 &{}+{}& {}^5 {\rm{C}}_4\, x^{5-4}\,(-1)^4\,(2y)^4 &{}+{}& {}^5 {\rm{C}}_5\, x^{5-5}\,(-1)^5\,(2y)^5&{}& {} &{} \\ {}={}&1 × x^5 × 1 × (2y)^0 &{}+{}& 5 × x^4 × -1 × (2y) &{}+{}& 10 × x^3 × 1 × (4y^2) &{}+{}& 10 × x^2 × -1 × (8y^3) &{}+{}& 5 × x^1 × 1 × (16y^4) &{}+{}& 1 × x^0 × -1 × (32y^5) \\ {}={}&x^5 &{}-{}& 10x^4 y &{}+{}& 40x^3 y^2 &{}-{}& 80 x^2 y^3 &{}+{}& 80 x y^4 &{}-{}& 32y^5 \\ \end{array}$

---

Case 2:
In this case, we investigate the expansion when the first quantity is 1. This can be written in 3 steps:
1. Let us put a = 1 and b = x. Then we get:
(a+b)ⁿ = (1+x)ⁿ
2. Expanding this using the general form, we get:
$(1+x)^n$
$\begin{array}{ll} {}={}&{}^n {\rm{C}}_0 \,1^n\,x^0 &{}+{}& {}^n {\rm{C}}_1\, 1^{n-1}\,x^1 &{}+{}& {}^n {\rm{C}}_2 \,1^{n-2}\,x^2 &{}+{}& {}^n {\rm{C}}_3\, 1^{n-3}\,x^3 &{}+{}& .~.~.~+~{}^n {\rm{C}}_n\, 1^0\,x^n&{}& {} &{} \\ {}={}&{}^n {\rm{C}}_0 \,1^n\,x^0 &{}+{}& {}^n {\rm{C}}_1\, 1^{n-1}\,x^1 &{}+{}& {}^n {\rm{C}}_2 \,1^{n-2}\,x^2 &{}+{}& {}^n {\rm{C}}_3\, 1^{n-3}\,x^3 &{}+{}& .~.~.~+~{}^n {\rm{C}}_n\, 1^0\,x^n&{}& {} &{} \\ \end{array}$
${}={}~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\, 1^{n-k}\;x^k}$
3. So we can write:
$(1+x)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;x^k}$
◼ Let us expand (1+2x)⁶ using this result. We get:
$(1+2x)^6~=~\sum\limits_{k\,=\,0}^{k\,=\,6}{{}^6 {\rm{C}}_k\;(2x)^k}$
• Once we write the sigma notation on the RHS, we will be able to write all the terms easily. We get:
$(1+2x)^6$
$\begin{array}{ll} {}={}&{}^6 {\rm{C}}_0 \,1^6\,(2x)^0 &{}+{}& {}^6 {\rm{C}}_1\,(2x)^1 &{}+{}& {}^6 {\rm{C}}_2\,(2x)^2 &{}+{}& {}^6 {\rm{C}}_3\,(2x)^3 &{}+{}& {}^6 {\rm{C}}_4\,(2x)^4 &{}+{}& {}^6 {\rm{C}}_5\,(2x)^5 &{}+{}& {}^6 {\rm{C}}_6\,(2x)^6&{}& {} &{} \\ {}={}&1 × (2x)^0 &{}+{}& 6 × (2x) &{}+{}& 15 × (4x^2) &{}+{}& 20 × (8x^3) &{}+{}& 15 × (16x^4) &{}+{}& 6 × (32x^5) &{}+{}& 1 × (64x^6) \\ {}={}&1 &{}+{}& 12x &{}+{}& 60 x^2 &{}+{}& 160 x^3 &{}+{}& 240 x^4 &{}+{}& 192 x^5 &{}+{}& 64x^6 \\ \end{array}$

◼ x = 1 is a special case coming under this category. We get:
$(1+x)^n~=~(1+1)^n~=~2^n~=~\sum\limits_{k\,=\,0}^{k\,=\,6}{{}^n {\rm{C}}_k\;x^k}~=~\sum\limits_{k\,=\,0}^{k\,=\,6}{{}^n {\rm{C}}_k\;1^k}~=~\sum\limits_{k\,=\,0}^{k\,=\,6}{{}^n {\rm{C}}_k}$
• Once we write the sigma notation on the RHS, we will be able to write all the terms easily. We get:
${}={}~{}^n {\rm{C}}_0 ~+~{}^n {\rm{C}}_1 ~+~{}^n {\rm{C}}_2 ~+~{}^n {\rm{C}}_3 ~+~.~.~.~+~{}^n {\rm{C}}_n$
• So we can write:
$2^n~=~{}^n {\rm{C}}_0 ~+~{}^n {\rm{C}}_1 ~+~{}^n {\rm{C}}_2 ~+~{}^n {\rm{C}}_3 ~+~.~.~.~+~{}^n {\rm{C}}_n$  

---

Case 3:
In this case, we investigate the expansion when the first quantity is 1 and the second quantity is -ve. This can be written in 3 steps:
1. Let us put a = 1 and b = -x. Then we get:
(a+b)ⁿ = (1-x)ⁿ = [1 + (-x)]ⁿ
2. Expanding this using the general form, we get:
$[1+(-x)]^n$
$\begin{array}{ll} {}={}&{}^n {\rm{C}}_0 \,1^n\,(-x)^0 &{}+{}& {}^n {\rm{C}}_1\, 1^{n-1}\,(-x)^1 &{}+{}& {}^n {\rm{C}}_2 \,1^{n-2}\,(-x)^2 &{}+{}& {}^n {\rm{C}}_3\, 1^{n-3}\,(-x)^3 &{}+{}& .~.~.~+~{}^n {\rm{C}}_n\, 1^0\,(-x)^n&{}& {} &{} \\ {}={}&{}^n {\rm{C}}_0 \,1^n\,(-1)^0\,x^0 &{}+{}& {}^n {\rm{C}}_1\, 1^{n-1}\,(-1)^1\,x^1 &{}+{}& {}^n {\rm{C}}_2 \,1^{n-2}\,(-1)^2\,x^2 &{}+{}& {}^n {\rm{C}}_3\, 1^{n-3}\,(-1)^3\,x^3 &{}+{}& .~.~.~+~{}^n {\rm{C}}_n\, 1^0\,(-1)^n\,x^n&{}& {} &{} \\ \end{array}$
${}={}~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k \, 1^{n-k}\;x^k}$
3. Based on the above result, we can write:
$(1-x)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k \;x^k}$

◼ Let us expand (1-2x)⁵ using this result. We get:
$(1-2x)^5~=~[1+(-2x)]^5~=~\sum\limits_{k\,=\,0}^{k\,=\,5}{{}^5 {\rm{C}}_k\;(-1)^k\;(-2x)^k}$
• Once we write the sigma notation on the RHS, we will be able to write all the terms easily. We get:
$(1-2x)^5$
$\begin{array}{ll} {}={}&{}^5 {\rm{C}}_0\,(-1)^0\,(2x)^0 &{}+{}& {}^5 {\rm{C}}_1\,(-1)^1\,(2x)^1 &{}+{}& {}^5 {\rm{C}}_2\,(-1)^2\,(2x)^2 &{}+{}& {}^5 {\rm{C}}_3\,(-1)^3\,(2x)^3 &{}+{}& {}^5 {\rm{C}}_4\,(-1)^4\,(2x)^4 &{}+{}& {}^5 {\rm{C}}_5\,(-1)^5\,(2x)^5&{}& {} &{} \\ {}={}&1 × 1 × (2x)^0 &{}+{}& 5 × -1 × (2x) &{}+{}& 10 × 1 × (4x^2) &{}+{}& 10 × -1 × (8x^3) &{}+{}& 5 × 1 × (16x^4) &{}+{}& 1 × -1 × (32x^5) \\ {}={}&1 &{}-{}& 10 x &{}+{}& 40 x^2 &{}-{}& 80 x^3 &{}+{}& 80 x^4 &{}-{}& 32 x^5 \\ \end{array}$

• We wrote: $(1-x)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k \;x^k}$
◼ x = 1 is a special case coming under this category. We get:
$(1-x)^n~=~(1-1)^n~=~0^n ~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k \;1^k} ~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k }$
• Once we write the sigma notation on the RHS, we will be able to write all the terms easily. We get:
${}={}~{}^n {\rm{C}}_0\;(-1)^0 ~+~{}^n {\rm{C}}_1\;(-1)^1 ~+~{}^n {\rm{C}}_2\;(-1)^2 ~+~{}^n {\rm{C}}_3 ~+~.~.~.~+~{}^n {\rm{C}}_n\;(-1)^n$
• So we can write:
$0^n~=~{}^n {\rm{C}}_0 ~-~{}^n {\rm{C}}_1 ~+~{}^n {\rm{C}}_2 ~-~{}^n {\rm{C}}_3 ~+~.~.~.~+~(-1)^n~ × ~{}^n {\rm{C}}_n$

---

In the next section we will see some solved examples.

Previous

Contents

Next

Copyright©2022 Higher secondary mathematics.blogspot.com