Mathematics Introduction , Binomial Theorem for Positive Integral Indices and Pascal's Triangle
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### Topics Covered

star Introduction
star Binomial Theorem for Positive Integral Indices
star Pascal’s Triangle

### Introduction

\color{green} ✍️ we have learnt how to find the squares and cubes of binomials like a + b and a – b. Using them, we could evaluate the numerical values of numbers like (98)^2 = (100 – 2)^2, (999)^3 = (1000 – 1)^3, etc.

\color{green} ✍️ However, for higher powers like (98)^5, (101)^6, etc., the calculations become difficult by using repeated multiplication. This difficulty was overcome by a theorem known as binomial theorem.

\color{green} ✍️ It gives an easier way to expand (a + b)^n, where n is an integer or a rational number .

### Binomial Theorem for Positive Integral Indices

Let us have a look at the following identities done earlier:
\color{green} ✍️ (a+ b)^0 = 1 , \ \ \ \ \ \ \ a + b ≠ 0
\color{green} ✍️ (a+ b)^1 = a + b
\color{green} ✍️ (a+ b)^2 = a^2 + 2ab + b^2
\color{green} ✍️ (a+ b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3
\color{green} ✍️ (a+ b)^4 = (a + b)^3 (a + b) = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)
we observe that
(i) The total number of terms in the expansion is one more than the index. For example, in the expansion of (a + b)^2 , number of terms is 3 whereas the index of (a + b)^2 is 2.

(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.

We now arrange the coefficients in these expansions as follows (Fig 8.1) :

\color { maroon} ® \color{maroon} ul (" REMEMBER")
● It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row for index 2.

● The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in the row for index 3 and so on.

● Also, 1 is present at the beginning and at the end of each row. This can be continued till any index of our interest.

### Pascal’s Triangle

\color{purple}ul(✓✓) \color{purple} (" DEFINITION ALERT")
\color{green} ✍️ The structure given in Fig 8.2 looks like a triangle with 1 at the top vertex and running down the two slanting sides. This array of numbers is known as "Pascal’s triangle," after the name of French mathematician Blaise Pascal. It is also known as Meru Prastara by Pingla.

Let us expand (2x + 3y)^5 by using Pascal’s triangle. The row for index 5 is
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ 5\ \ \ \ \ \ 10\ \ \ \ \ \ \ \ 10 \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ 1

\color{green} ✍️ (2x + 3y)^5 = (2x)^5 + 5(2x)^4 (3y) + 10(2x)^3 (3y)^2 +10 (2x)^2 (3y)^3 + 5(2x)(3y)^4 +(3y)^5

 \ \ \ \ \ \ \ \ \ \ \ \ \ = 32x^5 + 240x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5.

Now, if we want to find the expansion of (2x + 3y)^12, we are first required to get the row for index 12.
● This can be done by writing all the rows of the Pascal’s triangle till index 12. This is a slightly lengthy process.

● The process, as you observe, will become more difficult, if we need the expansions involving still larger powers.

● For this, we make use of the concept of combinations studied earlier to rewrite the numbers in the Pascal’s triangle.

\color { maroon} ® \color{maroon} ul (" REMEMBER") We know that ""^nC_r= (n!)/(r!(n-r)!)  \ \ \ 0 ≤ r ≤ n and n is a non-negative integer. Also, ""^nC_0 = 1 = ""^nC_n

\color{blue} \mathbf( "The Pascal’s triangle can now be rewritten as (Fig 8.3)")

● Observing this pattern, we can now write the row of the Pascal’s triangle for any index without writing the earlier rows.

For example, for the index 7 the row would be
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ""^7C_0\ \ \ \ \ ""^7C_1\ \ \ \ \ \ ""^7C_2 \ \ \ \ \ \ \ \ ""^7C_3 \ \ \ \ \ \ \ ""^7C_4 \ \ \ \ \ \ \ \ ""^7C_5 \ \ \ \ \ \ \ \ \ \ ""^7C_6 \ \ \ \ \ \ \ \ \""^7C_7.

\color{green} ✍️(a + b)^7 = ""^7C_0 a^7 + ""^7C_1a^6b + ""^7C_2a^5b^2 + ""^7C_3a^4b^3 + ""^7C_4a^3b^4 + ""^7C_5a^2b^5 + ""^7C_6ab^6 + ""^7C_7b^7

We are now in a position to write the expansion of a binomial to any positive integral index.

### Binomial theorem for any positive integer n

\color{green} ✍️ color(blue)((a + b)^n = ""^nC_0a^n + ""^nC_1a^(n–1)b + ""^nC_2a^(n–2) b^2 + ......+ ""^nC_(n – 1)a.b^(n–1) + ""^nC_nb^n)

The notation sum_(k=0)^n ""^nC_ka^(n-k)b^k stands for
""^nC_0a^n + ""^nC_1a^(n–1)b + ""^nC_2a^(n–2) b^2 + ..........+""^nC_ra^(n–r)b^r....+ ""^nC_(n – 1)a.b^(n–1) + ""^nC_na^(n–n)b^n

Hence the theorem can also be stated as color(red)((a+b)^n= sum_(k=0)^n ""^nC_ka^(n-k)b^k)

\color{green} ✍️ The coefficients ""^nC_r occurring in the binomial theorem are known as binomial coefficients.

\color{green} ✍️ There are (n+1) terms in the expansion of (a+b)^n, i.e., one more than the index.

\color{green} ✍️ In the successive terms of the expansion the index of a goes on decreasing by unity. It is n in the first term, (n–1) in the second term, and so on ending with zero in the last term. At the same time the index of b increases by unity, starting with zero in the first term, 1 in the second and so on ending with n in the last term.

\color{green} ✍️ In the expansion of (a+b)^n, the sum of the indices of a and b is n + 0 = n in the first term, (n – 1) + 1 = n in the second term and so on 0 + n = n in the last term. Thus, it can be seen that the sum of the indices of a and b is n in every term of the expansion.

### Some special cases

In the expansion of

\color{green} ✍️ color(green)((a + b)^n = ""^nC_0a^n + ""^nC_1a^(n–1)b + ""^nC_2a^(n–2) b^2 + ......+ ""^nC_(n – 1)a.b^(n–1) + ""^nC_nb^n)

\color{green} ✍️ (i) Taking a = x and b = – y, we obtain

color{blue} ((x – y)^n = [x + (–y)]^n)

\ \ \ \ \ \ \ \ \ \ \ \= ""^nC_0x^n + ""^nC_1x^(n – 1)(–y) + ""^nC_2x^(n–2)(–y)^2 + ""^nC_3x^(n–3)(–y)^3 + ......... + ""^nC_n (–y)^n

 \ \ \ \ \ \ \ \ \ \ \ \ \= ""^nC_0x^n – ""^nC_1x^(n – 1)y + ""^nC_2x^(n – 2)y^2 – ""^nC_3x^(n – 3)y^3 + ......... + (–1)^n \ \ ""^nC_n y^n

Thus  \ \ \ \ \ (x–y)^n = ""^nC_0x^n –""^nC_1x^(n – 1) y + ""^nC_2x^(n – 2) y^2 + ........ + (–1)^n \ \ ""^nC_n y^n

Using this, we have
\color{green} ( ✍️(x–2y)^5 )= ""^5C_0x^5 – ""^5C_1x^4 (2y) + ""^5C_2x^3 (2y)^2 – ""^5C_3x^2 (2y)^3 + ""^5C_4 x(2y)^4 – ""^5C_5(2y)^5

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = x^5 –10x^4y + 40x^3y^2 – 80x^2y^3 + 80xy^4 – 32y^5.

\color{green} ✍️ (ii) Taking a = 1, \ \ b = x, we obtain

color{blue} ((1 + x)^n) = ""^nC_0(1)^n + ""^nC_1(1)^(n – 1)x + ""^nC_2(1)^(n – 2) x^2 + ........ + ""^nC_nx^n

 \ \ \ \ \ \ \ \ \ \ \ \ \ = ""^nC_0 + ""^nC_1x + ""^nC_2x^2 + ""^nC_3x^3 + ... + ""^nC_nx^n

Thus (1 + x)^n = ""^nC_0 + ""^nC_1x + ""^nC_2x^2 + ""^nC_3x^3 + ........... + ""^nC_nx^n

In particular, for x = 1, we have

\color { maroon} ® \color{maroon} ul (" REMEMBER")  \ \ \ \ \ \ \ \ \ \ \color{blue} (2^n = ""^nC_0 + ""^nC_1 + ""^nC_2 + ........ + ""^nC_n).

(iii) Taking a = 1, \ \ b = – x, we obtain

(1– x)^n = ""^nC_0 – ""^nC_1x + ""^nC_2x^2 – ............. + (– 1)^n \ \ ""^nC_nx^n

In particular, for x = 1, we get

\color { maroon} ® \color{maroon} ul (" REMEMBER") \ \ \ \ \ \ \ \ \ \ \ \color{blue}(0 = ""^nC_0 – ""^nC_1 + ""^nC_2 – ......... + (–1)^n \ \ ""^nC_n)
Q 3181156027

Expand ( x^2+3/x)^4 , x ne 0

Solution:

By using binomial theorem, we have

(x^2+3/x)^4 = text()^4C_0 (x^2)^4+text()^4C_1 (x^2)^3 (3/x) +text()^4C_2(x^2)^2 (3/x)^2+text()^4C_3(x^2)(3/x)^3+text()^4C_4(3/x)^4

 = x^8+4x^6 . 3/x + 6 . x^4 . 9/x^2 +4. x^2 27/x^3 +81/x^4

 = x^8+12x^5+54x^2+108/x +81/x^4
Q 3141256123

Compute (98)^5.

Solution:

We express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
Write 98 = 100 – 2
Therefore, (98)^5 = (100 – 2)^5

= text()^5C_0 (100)^5 - text()^5C_1 (100)^4 .2 +text()^5C_2 (100)^3 2^2 - text()^5C_3 (100)^2 (2)^3+text()^5C_4(100)(2)^4 - text()^5C_5(2)^5

= 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32

= 10040008000 – 1000800032 = 9039207968.
Q 3131356222

Which is larger (1.01)^(1000000) or 10,000?

Solution:

Splitting 1.01 and using binomial theorem to write the first few terms we have

(1.01)^(1000000) = (1+0.01)^(1000000)

= text()^(1000000)C_0+text()^(1000000)C_1 (0.01) + other positive terms

= 1 + 1000000 × 0.01 + other positive terms

= 1 + 10000 + other positive terms

> 10000

Hence (1.01)^(1000000) > 10000
Q 3131456322

Using binomial theorem, prove that 6^n–5n always leaves remainder 1 when divided by 25.

Solution:

For two numbers a and b if we can find numbers q and r such that a = bq + r, then we say that b divides a with q as quotient and r as remainder. Thus, in order to show that 6^n – 5n leaves remainder 1 when divided by 25, we prove that 6^n – 5n = 25k + 1, where k is some natural number.

We have

(1+a)^n = text()^nC_0 +text()^nC_1 a+text()^nC_2 a^2+...........+ text()^nC_na^n

(1+5)^n = text()^nC_0+text()^nC_1 5 +text()^nC_2 5^2 + ................+ text()^nC_n 5^n

(6)^n = 1+5n+5^2 . text()^nC_2+5^3 . text()^nC_3 +...........+ 5^n

6^n -5n = 1+5^2 ( text()^nC_2+text()^nC_3 5+.........+ 5^(n-2) )

or 6^n -5n = 1+25 ( text()^nC_2+5 . text()^nC_3 +.......+ 5^(n-2))

or 6^n -5n = 25k+1 where k = text()^nC_2 +5 . text()^nC_3 + ..............+ 5^(n-2)

This shows that when divided by 25, 6^n – 5n leaves remainder 1.