**A Tutorial on the Binomial Theorem and Binomial Coefficients: Notes, Figures and Problems with Solutions**

Target Audience: These notes on Atomic Structure are meant for college freshmen, or high school students in Grades 11 or 12.

They
might be of use to Indian students preparing for the ISC or CBSE Class
11 and 12 Examinations, IIT JEE (main and advanced), AIEEE; students
from across the world preparing for their A Level Examinations, IB
(International Baccalaureate) or AP Mathematics.

**This compilation of notes has been prepared by Ayushi Patel of IIT Gandhinagar.**

**Complete Tutorial with Problems, Figures and Solutions : **

**Topics that are covered in theory and tutorials**

**You may either refer to the PDF above (preferable) or go through the notes below. They are the same, however, the former is more presentable and easier to comprehend.**

1)Binomial expansion

2)General term of Binomial expansion

3)Coefficient of xr in a binomial expansion.

4)Middle term of (x+y)n when n is odd.

5)Middle term of (x+y)n when n is even.

6)Important formulae including reduction formula.

7)Greatest binomial coefficient and term.

8)(1+x)-n expansion

9)Check inequality between two real numbers using binomial formulae.

10)Remainder when divided by a number

11)Finding terms independent of x

12)Pascals triangle.

13)Finding total number of terms.

14)Finding given real is an integer or not.

15)Checking divisibilty.

Binomial theorem and expansion

1) Binomial expansion

(x+y)n = nC0 xn + nC1 xn-1 y + nC2 xn-2 y2 +....... nCn yn where x ,y both are real numbers.

(1+x)n = nC0 xn + nC1 xn-1 + nC2 xn-2 +....... nCn

We can find expansion for 2n by putting x = 1 in above equation .

2n = nC0 + nC1 + nC2 +....... nCn

2n-1= nC0 + nC2 +............... = nC1 + nC3 +.......

We can see in every expansion the total number of terms for any expansion with degree n is (n+1)

2) General term for binomial expansion:

It can be written in 2 ways and both ways give the same result. .

Tr +1 = nCr xn-r yr

= nCr yn-r xr

3)For finding coefficient of xr search the coeficient of xr in the binomial expansion of the given expression (We will see one example for this in the tutorials .

4)Middle term of (x+y)n

i)When n is even number :

n+1 is the total number of terms which is odd,therefore we will have one middle term which will be (n/2 + 1)th term .

ii)When n is odd number :

n+1 is the total number of terms which is even,therfore we will have two middle terms that will be (n+1 )th/2 and ( n+3 )th/2 term.

4)Important formulae:

Reduction formulae: nCr + nCr-1 = n+1Cr

r nCr = n n-1Cr-1

Above formulae can also be written as

nCr (1+n) = (1+r) n+1Cr+1

Tr +1 nCr xn-r yr

____ = ________

Tr nCr-1 xn-r+1 yr-1

6)Greatest binomial coefficient and term:

Greatest binomial coefficient is the coefficient of the middle term.

But,greatest binomial term is not necessarily the middle term,it occurs where

Tr +1 > Tr

(1+x)-n =1 + nC1 x1 + nC2 x2 +.......infinite terms.

Pascal’s triangle :

(x+y)0 = 1

(x+y) =1x + 1y

(x+y)2 = x2 + 2xy+ y2 .

(x+y)3 =x3 + 3x2y + 3xy2 + y3.

Pascals Triangle:

1

1 1

1 2 1

1 3 3 1

In this way Pascals triangle is formed.

Expansion of (1+x)n when n is a real number .

(1+x)n = 1 + nx + n(n-1) x2 + n(n-1)(n-2) +…………∞

_________ _________

2 ! 3!

Tutoral and Solved Problems:

**Questions: (Solutions are given at the end)**

Question 1

Find the term which is independent of x in the expression {(x/2)1/2 + 21/2/x3/2 }8 .

Question 2)

Find if 6n – 10n – 6 is divisible 5 or not .If it is not divisible, find the remainder if its not completely divisible.

Question 3)

Find which among the two is greater 8040 + 7940 or 8140 .

Question 4)

Find the remainder when 2202 is divided by 15.

Question 5)

Find the total number of terms for the given expression (x + 72)200 + (x-72)200 .

Question 6)

Find whether { ((5)1/2 + 1)150 + ((5)1/2 - 1)150 } is irrational number or a rational number.

Question 7)

For the given expression find the greatest binomial term (1-5x)8 ,take x=1 .

Question 8)

Express the following as a combination of 2 numbers

12C4 + 14C6 + 12C3 + 13C5 + 15C7 .

Question 9)

Prove that x6 - 531441 is divisible by (x-9) using binomial theorem.

Question 10)

Write the expansion of (1 – (7)1/3 )√5 .

Question 11)

Find the coefficient of middle term for the expansion (3-7x1/5)19 .

Question 12)

Find the coefficient of x4 in the expansion (1+x3)4 (1+x)5 .

Question 13)

If the 4th term is negative of 5th term , and the ratio of q : p is 5:1 in the expression (p+q)n .Find the value of n.

Question 14)

Find the sum of the following

5C0 + 3.5C1 + 10.5C2 +28. 5C3 + 82.5C4 + 244.5C5

Question 15)

Express nC0 .nCn + nC1 .nCn + nC2 .nCn-2 + ....... nC0 .nCn

as a combination of two numbers.

Tutorials with Solutions:

Question 1)

Find the term which is independent of x in the expression {(x/2)1/2 + 21/2/x3/2 }8 .

Solution 1)

For finding term independent of x,write firstly the general term for the expression.

Tr+1 = nCr xn-r yr for binomial expression of type (x+y)n .

Tr+1 = 8Cr ((x/2)1/2)8-r (21/2/x3/2 )r

Tr+1 = 8Cr ( x4-r/2 . x-3r/2 ). 2(-4+r/2+r/2)

Tr+1 = constant . x(4-4r/2)

Since we want term independent of x

For 4-4r/2 =0 which gives r=2 .

Therefore the term independent of x is

Tr+1 = 8Cr ((x/2)1/2)8-2 (21/2/x3/2 )2 .

= 8Cr .2-2 . 2

Therefore,term independent of x is 8Cr /2 .

Question 2)

Find if 6n – 10n – 6 is divisible 5 or not .If it is not divisible, find the remainder if its not completely divisible.

Solution 2)

If we are able to proof that 6n – 10n – 6 =5k,then we can say that it is divisible by 5.

Using formula, (1+x)n = nC0 xn + nC1 xn-1 + nC2 xn-2 +....... nCn

We can write

(1+5)n = nC0 + nC1 5 + nC2 52 +……….. nCn 5n

= 1+ 5n + 52 (nC2 + ……. nCn 5n-2 )

6n -5n -1 = 52 (nC2 + ……. nCn 5n-2 )

Subtraction 5(n-1) from both sides we get

6n – 10n – 6 = 5{5(nC2 + ……. nCn 5n-2 ) –n- 1}

= 5k

Since 6n – 10n – 6 can be expressed as 5k therefore,remainder is 0 and 6n – 10n – 6 is divisible by 5.

Question 3)

Find which among the two is greater 8040 + 7940 or 8140 .

Solution 3)

Let us find

8140 - 7940 = (1+80)40 – (1-79)40

= { 1+ 40C1 .80 + 40C2 .802 + ……..40C40 . 8040 } - { 1- 40C1 .80 + 40C2 .802 +

…..40C40 . 8040

In the above expression,terms with even power will cancel out and terms with odd power will remain.

8140 - 7940 = 2 . 40C1 .80 +……………….+ 2 . 40C39 .8039

= 2 . 40C1 .80 +……………….+ 2 .40.8039

= 2 . 40C1 .80 +……………….+ 8040

Therefore , 8140 - 7940 > 8040

8140 > 7940 + 8040

Question 4)

Find the remainder when 2202 is divided by 15.

Solution 4)

2202 = 4101 can be written as 4.4100 = 4.(44 )25 .

= 4.1625 .

= 4.(1+15)25 .

4.(1+15)25 = 4.(1+ 25C1.15 + 25C2 .152 + ……. 25C25 .1525

= 4+ 15(25C1 + 25C2 .15 + ……. 25C25 .1524 )

Therefore,the remainder is 4.

Question 5)

Find the total number of terms for the given expression (x + 72)200 + (x-72)200 .

Solution 5)

By using formulae from the theory we can write

(x+y)n = nC0 xn + nC1 xn-1 .y + nC2 xn-2 .y2 +....... nCn yn .

(x + 72)200 = 200C0 x200 + 200C1 x200-1 .72 + 200C2 x200-2 .722 +....... 200C200 .72200 .

(x - 72)200 = 200C0 x200 - 200C1 x200-1 .72 + 200C2 x200-2 .722 +....... 200C200 .72200 .

Both of the above equations has n+1 terms ,adding these two ,we can see that terms with odd power of x will cancel out and even terms will remain ,total number of terms for ( x + 72)n is 200+1 = 201 out of which 101 are even terms and 100 are odd terms so 100 terms after adding these two equations will cancel out and the rest 101 will add up according to their powers.Therefore,the total number of terms will be 101.

Question 6)

Find whether { ((5)1/2 + 1)150 + ((5)1/2 - 1)150 } is irrational number or a rational number.

Solution 6)

Using the formula

(1+x)n = nC0 xn + nC1 xn-1 + nC2 xn-2 +....... nCn

(1+(5)1/2 )150 = 1 + 150C1 .(5)1/2 + 150C2 . 5 +……… + 150C150 .575 .

(1-(5)1/2 )150 = 1 - 150C1 .(5)1/2 + 150C2 . 5 +……… + 150C150 .575 .

Here also,after adding the above two equations, the terms with odd power of x i.e 51/2 will cancel out and terms with even power of 51/2 will remain and terms with even power of 51/2 will also be rational(since when an even number is divided by 2,an integer is obtained and 5 to the power an integer is a rational number.

Since,only rational number is remaining after addition ,

Therefore, { ((5)1/2 + 1)150 + ((5)1/2 - 1)150 } is a rational number .

Question 7)

For the given expression find the greatest binomial term (1-5x)8 ,take x=1 .

Solution 7)

As we studied in the theory,greatest binomial term occurs when

Tr+1 > Tr i.e when ratio of these is greter than 1.Writing Tr+1

Tr+1 = 8Cr (1)8-r.(-5x)r

Tr+1 > 1

___

Tr

n-r+1 |y| > 1 for (x+y)n

_________

r |x|

(8-r+1) x 5x1 > 1

___________

r

r<45/6

r<7.5

r=7

Therefore,the greatest binomial term is T8 =8C7 (1)8-7.(-5x)7 .

Question 8)

Express the following as a combination of 2 numbers

12C4 + 14C6 + 12C3 + 13C5 + 15C7 .

Solution 8)

We can do this using reduction formula,

nCr + nCr-1 = n+1Cr

Using reduction formula for 12C3 and 12C4 we get 13C4 .

Using reduction formula for 13C4 and 13C5 we get 14C5 .

Using reduction formula for 14C5 and 14C6 we get 15C6 .

Using reduction formula for 15C6 and 15C7 we get 16C7 .

Question 9)

Prove that x6 - 531441 is divisible by (x-9) using binomial theorem.

Solution 9)

531441 = 312 = 96 .

Therefore ,x6 - 531441 = x6 - 96 .

x6 can be written as (x -9 + 9)6 which can further be written using binomial expansion as

(x -9 + 9)6 = 6C0 96 (x-9)0 + 6C1 95 .(x-9)1 + 6C2 94 .(x-9)2 +....... 6C6 .(x-9)6 .

(x -9 + 9)6 - 96 = 96 + 6C1 95 .(x-9)1 + 6C2 94 .(x-9)2 +....... 6C6 .(x-9)6 - 96 .

= (x-9) {6C1 95 .(x-9)0 + 6C2 94 .(x-9)1 +....... 6C6 .(x-9)5 }

Therefore ,x6 - 531441 is divisible by (x-9).

Question 10)

Write the expansion of (1 – (7)1/3 )√5 .

Solution 10)

Using

expansion of (1+x)n when n is a real number .

(1+x)n = 1 + nx + n(n-1) x2 + n(n-1)(n-2) +…………∞

_________ _________

2 ! 3!

We can write

(1 – (7)1/3 )√5 = 1 + √5 (–1)(7)√3 - √5 (√5 -1)(7)3/2 +……………………….∞

____________

2!

Question 11)

Find the coefficient of middle term for the expansion (3-7x1/5)19 .

Solution 11)

Since ,n is odd we will have 2 middle terms n=19 here ,

So middle term = (n+1)/2 th and (n+3)/2th term

(n+1)/2 = 10th term

(n+3)/2 = 11th term

Tr+1 = nCr xn-r yr for binomial expression of type (x+y)n .

r=9 for 10th term and r=10 for 11th term.

T10 = 19C9 310 .(-7x1/5)9 .

T11 = 19C10 39 .(-7x1/5)10 .

Question 12)

Find the coefficient of x4 in the expansion (1+x3)4 (1+x)5 .

Solution 12)

(1+x)n = nC0 + nC1 x1 + nC2 x2 + ....... xn.nCn

(1+x)5 = 5C0 + 5C1 x1 + 5C2 x2 + 5C3 x3 + 5C4 x4 + 5C5 x5 ----------------------------------------- 1)

(x+y)n = nC0 xn + nC1 xn-1 .y + nC2 xn-2 .y2 +....... nCn yn .

(1+x3 )4 = 4C0 + 4C1 ( x3)1 + nC2 (x3)2 +....... ---------------------------------2)

Multiplying equation 1) and 2) we get coefficient of x4 as 5C4.4C0 + 5C1. 4C1

Question 13)

If the 4th term is negative of 5th term , and the ratio of q : p is 5:1 in the expression (p+q)n .Find the value of n.

Solution 13)

Tr+1 = nCr xn-r yr for binomial expression of type (x+y)n .

For 4th term r=3 in above equation and for 5th term r=4 in above equation.

T3+1 = nCr3 pn-3 q3 for given binomial expression _______________1).

T4+1 = nC4 pn-4 q4 for given binomial expression ___________________2) .

Dividing equation 2) by 1) we get,

T5 = nC4 pn-4 q4

__ __________ = -1 (According to the question )

T4 = nC3 pn-3 q3

and also putting q/p =5 in above equation,we get

q = (n-3)!

__ _____ = 5

p (n-4)!

This gives n=8.

Question 14)

Find the sum of the following

5C0 + 3.5C1 + 10.5C2 +28. 5C3 + 82.5C4 + 244.5C5

Solution 14)

5C0 + 3.5C1 + 10.5C2 +28. 5C3 + 82.5C4 + 244.5C5

Above expression is the sum of two expressions and they are,

25 = 5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5

45 = 5C0 + 35C1 + 32.5C2 +33. 5C3 + 34.5C4 +35.5C5

Therefore,the sum of the above expression is 25 + 45 .

Question 15)

Express nC0 .nCn + nC1 .nCn + nC2 .nCn-2 + ....... nC0 .nCn

as a combination of two numbers.

Solution 15)

nC0 .nCn + nC1 .nCn-1 + nC2 .nCn-2 + ....... nC0 .nCn

Let us consider above expression as equation 1)

(1+x)n = nC0 + nC1 x1 + nC2 x2 + ....... xn.nCn

(1+x)n = nC0 + nC1 x1 + nC2 x2 + ....... xn.nCn

Multiplying the above equations, we get

(1+x)2n = ( nC0 + nC1 x1 + nC2 x2 + ....... xn.nCn ). (nC0 + nC1 x1 + nC2 x2 + ....... xn.nCn )

We can see that equation 1) is actually coefficient of xn in the RHS of above expression.

Therefore the value of equation 1) from LHS should also be coefficient of xn

which is equal to 2nCn .

Hence,equation 1) can be expressed as 2nCn

References:

NCERT text book

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