Mathematics Introduction , Binomial Theorem for Positive Integral Indices and Pascal's Triangle
Please Wait... While Loading Full Video

Class 11 Chapter 8 - BINOMIAL THEOREM

Introduction , Binomial Theorem for Positive Integral Indices and Pascal's Triangle

Click for Only Video

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`.
 
SiteLock