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.

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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:
$(a+b)^n~=~{}^n {\rm{C}}_0 a^n~+~{}^n {\rm{C}}_1 a^{n-1} b~+~{}^n {\rm{C}}_2 a^{n-2} b^2~+~{}^n {\rm{C}}_3 a^{n-3} b^3~+~.~.~.~+~{}^n {\rm{C}}_{n-1} a b^{n-1}~+~{}^n {\rm{C}}_n b^n$

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• 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}$

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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.

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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}$

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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:
$2^n~=~{}^n {\rm{C}}_0
~+~{}^n {\rm{C}}_1
~+~{}^n {\rm{C}}_2
~+~{}^n {\rm{C}}_3
~+~.~.~.~+~{}^n {\rm{C}}_n$  

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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:
$0^n~=~{}^n {\rm{C}}_0
~-~{}^n {\rm{C}}_1
~+~{}^n {\rm{C}}_2
~-~{}^n {\rm{C}}_3
~+~.~.~.~+~(-1)^n~ × ~{}^n {\rm{C}}_n$

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In the next section we will see some solved examples.

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Chapter 8 - Binomial Theorem

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 5^th 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.

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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.

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:

Fig.8.2

• Similarly,
    ♦ 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.

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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

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• 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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