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( )
n
x a+
( )
1
x a x a+ = +
A binomial is a polynomial with two terms such as x + a. Often we
need to raise a binomial to a power. In this section we'll explore a
way to do just that without lengthy multiplication.
( )
2 2 2
2x a x ax a+ = + +
( )
3 3 2 2 3
3 3x a x ax a x a+ = + + +
( )
4 4 3 2 2 3 4
4 6 4x a x ax a x a x a+ = + + + +
Can you see a
pattern?
Can you make a
guess what the next
one would be?
( )
5
x a+ = 5 4 2 3 3 2 4 5
__ __ __ __x ax a x a x a x a+ + + + +
We can easily see the pattern on the x's and the a's. But what
about the coefficients? Make a guess and then as we go we'll
see how you did.
( )
0
1x a+ =
Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
This is called Pascal's Triangle and would give us the
coefficients for a binomial expansion of any power if we
extended it far enough.
This is good for
lower powers but
could get very
large. We will
introduce some
notation to help
us and
generalise the
coefficients with
a formula based
on what was
observed here.
!
!!
!The Factorial Symbol
0! = 1 1! = 1
n! = n(n-1) · . . . · 3 · 2 · 1
n must be an integer greater than or equal to 2
What this says is if you have a positive integer followed by
the factorial symbol you multiply the integer by each integer
less than it until you get down to 1.
6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
Your calculator can compute factorials. The !
symbol is under the "math" menu and then "prob".
The binomial theorem class 11 maths
General
Term
The binomial theorem class 11 maths
Here is the expansion of (x + a)12
…and the 5th term matches the term we obtained!
In this expansion, observe the following:
•Powers on a and x add up to power on binomial
•a's increase in power as x's decrease in power from
term to term.
•Powers on a are one less than the term number
•Symmetry of coefficients (i.e. 2nd term and 2nd to last term
have same coefficients, 3rd & 3rd to last etc.) so once you've
reached the middle, you can copy by symmetry rather
than compute coefficients.
Let's use what we've learned to expand (2x - 3y)6
First let's write out the expansion of the general (x + a)6
and
then we'll substitute.
( )
6 6 5 2 4 3 3 4 2 5 6
__ __ __ __ __x a x ax a x a x a x a x a+ = + + + + + +
these will be the same
these will be the same
Let's find the
coefficient for the
second term.
6 6! 6 5!
6
1 1!5! 5!
  ×
= = = ÷
 
Let's confirm that
this is also the
coefficient of the
2nd to last term.
6 6! 6 5!
6
5 5!1! 5!
  ×
= = = ÷
 
6 6
Let's find the
coefficient for the
third term.
6 6! 6 5 4!
15
2 2!4! 2 4!
  × ×
= = = ÷
× 
15
This will
also be the
coefficient of
the 3rd to
last term.
15
Now we'll find
the coefficient of
the 4th term
6 6! 6 5 4 3!
20
3 3!3! 3 2 3!
  × × ×
= = = ÷
× × 
20
Now we'll apply this formula to our specific binomial.
Instead of x
we have 2x
Instead of a
we have -3y
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
6 6 5 2 4
3 3 4 2 5 6
6 15
20
2 2 2 2
2 215
3 3 3
3 3 3 36 2
y y y
y y y
x x x
x x y
x
x
= + + +
+ + +
− − −
− − − −
6 5 4 2 3 3
2 4 5 6
64 576 2160 4320
4860 2916 729
x x y x y x y
x y xy y
= − + −
+ − +
Name:- Pramod Mehra
Class:- XI A
Roll no.:-
School:- KV (AFS) Uttarlai,
Barmer
The binomial theorem class 11 maths

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The binomial theorem class 11 maths

  • 2. ( ) 1 x a x a+ = + A binomial is a polynomial with two terms such as x + a. Often we need to raise a binomial to a power. In this section we'll explore a way to do just that without lengthy multiplication. ( ) 2 2 2 2x a x ax a+ = + + ( ) 3 3 2 2 3 3 3x a x ax a x a+ = + + + ( ) 4 4 3 2 2 3 4 4 6 4x a x ax a x a x a+ = + + + + Can you see a pattern? Can you make a guess what the next one would be? ( ) 5 x a+ = 5 4 2 3 3 2 4 5 __ __ __ __x ax a x a x a x a+ + + + + We can easily see the pattern on the x's and the a's. But what about the coefficients? Make a guess and then as we go we'll see how you did. ( ) 0 1x a+ =
  • 4. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 This is called Pascal's Triangle and would give us the coefficients for a binomial expansion of any power if we extended it far enough. This is good for lower powers but could get very large. We will introduce some notation to help us and generalise the coefficients with a formula based on what was observed here.
  • 5. ! !! !The Factorial Symbol 0! = 1 1! = 1 n! = n(n-1) · . . . · 3 · 2 · 1 n must be an integer greater than or equal to 2 What this says is if you have a positive integer followed by the factorial symbol you multiply the integer by each integer less than it until you get down to 1. 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720 Your calculator can compute factorials. The ! symbol is under the "math" menu and then "prob".
  • 9. Here is the expansion of (x + a)12 …and the 5th term matches the term we obtained! In this expansion, observe the following: •Powers on a and x add up to power on binomial •a's increase in power as x's decrease in power from term to term. •Powers on a are one less than the term number •Symmetry of coefficients (i.e. 2nd term and 2nd to last term have same coefficients, 3rd & 3rd to last etc.) so once you've reached the middle, you can copy by symmetry rather than compute coefficients.
  • 10. Let's use what we've learned to expand (2x - 3y)6 First let's write out the expansion of the general (x + a)6 and then we'll substitute. ( ) 6 6 5 2 4 3 3 4 2 5 6 __ __ __ __ __x a x ax a x a x a x a x a+ = + + + + + + these will be the same these will be the same Let's find the coefficient for the second term. 6 6! 6 5! 6 1 1!5! 5!   × = = = ÷   Let's confirm that this is also the coefficient of the 2nd to last term. 6 6! 6 5! 6 5 5!1! 5!   × = = = ÷   6 6 Let's find the coefficient for the third term. 6 6! 6 5 4! 15 2 2!4! 2 4!   × × = = = ÷ ×  15 This will also be the coefficient of the 3rd to last term. 15 Now we'll find the coefficient of the 4th term 6 6! 6 5 4 3! 20 3 3!3! 3 2 3!   × × × = = = ÷ × ×  20 Now we'll apply this formula to our specific binomial. Instead of x we have 2x Instead of a we have -3y ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6 6 5 2 4 3 3 4 2 5 6 6 15 20 2 2 2 2 2 215 3 3 3 3 3 3 36 2 y y y y y y x x x x x y x x = + + + + + + − − − − − − − 6 5 4 2 3 3 2 4 5 6 64 576 2160 4320 4860 2916 729 x x y x y x y x y xy y = − + − + − +
  • 11. Name:- Pramod Mehra Class:- XI A Roll no.:- School:- KV (AFS) Uttarlai, Barmer