BINOMIAL EXPANSION

Expansion

(a + x)¹  = a + x

(a + x)² = a²+2ax+x²

(a + x)³ = a³+3a²x+3ax²+x³

(a + x)⁴ = a⁴+4a³x+6a²x²+4ax³+x⁴

(a + x)⁵ = a⁵+5a⁴x+10a³x²+10a²x³+5ax⁴+x⁵

(a + x)ⁿ = aⁿ+b₁a^{n – 1}x + b₂a^{n– 2}x²+b₃a^{n –  3}x³+  ------ +xⁿ

Note the following points in this expansion

1.                  The powers of a decreases by one while the powers of x increases by one in each term

2.                  The sum of the powers of a and x in each term is equal to the power of the binomial

3.                  There are (n+1) terms in each binomial, where n is the power of the binomial

4.                  The coefficients of both the first and last term are each one.

5.                  The powers of the first and the last term are each equal to n (n is the power of the binomial

6.                  With power arranged as shown below, the coefficient for  a pattern as shown below

BINOMIAL                              COEFFICIENT 

(a + x)¹                                                1       1

(a + x)²                                              1      2       1

(a + x)³                                      1      3            3          1

(a + x)⁴                                  1          4            6          4          1         

(a + x)⁵                               1             5            10        10        5      1

(a + x)⁶                        1          6      15            20        15        6            1

            The display of coefficient shown above is PASCAL’S TRIANGLE and it is used in determining the coefficient of terms in a binomial expression.  Each line of coefficient is obtained from the line of coefficient immediately above it.

e.g. (a+x)⁷

              1            6        15        20        15        6          1

           

Example

   Using Pascal’s is triangle, expand (a–x)⁴

Solution   

 From Pascal’s triangle, the coefficient one 1,4,6,4,1

(a–x)⁴ = [a+(–4)]⁴ = a⁴+4a³(–x)+6a²(–x)²+4a(–x)³+(–x)⁴

                              = a⁴ – 4a³x + 6a²x² – 4ax³ + x⁴

Example

   Using Pascal’s triangle, expand and simplify completely (3x+5y)⁴

Solution

From Pascal’s triangle, the coefficients are 1,4,6,4,1

(3x+5x)⁴ = (3x)⁴+4(3x)³(5y)+6(3x)²(5y)²+4(3x)(5y)³+(5y)⁴

(3x+5y)⁴ = 81x⁴ + 540x³y + 1350x²y²+1500xy³+625y⁴

Example

    Using Pascal’s triangle, expand (4x – 5y)⁵

Solution

 The coefficient are 1, 5, 10, 10, 5, 1

(4x – 5y)⁵ = [4x+(–5y)]⁵

                 = (4x)⁵ + 5(4x)⁴ (–5y) + 10(4x)³ (–5y)² + 10(4x)² (–5x)³ + 5(4x) (–5y)⁴ + (–5y)⁵

                 = 1024x⁵ – 6400x⁴y + 16000x³y² –20000x²y³+12500xy⁴ – 3125y⁵

Example

    Using Pascal’s triangle, simplify (1.01)⁴

Solution

(1.01)⁴ = (1+0.01)⁴

from Pascal’s triangle, the coefficients are 1,4,6,4,1

= 1⁴ + 4(1)³ (0.01) + 6(1)² (0.01)² + 4(1) (0.01)³ + (0.01)⁴

= 1+ 0.04 + 0.0006 + 0.000004 + 0.00000001

= 1.04060401

BINOMIAL THEOREM

For any number x, y and natural number the following hold true

(x+y)ⁿ = xⁿ + ⁿc₁x^n–1y¹ +  ⁿc₂x^n–2y² +  ⁿc₃x^n–3y³+ – – – + ⁿc_rx^n–ry^r + – –+ y^r

where ⁿc_r=

   

It can be proved that the binomial expansion holds true for positive, negative, integral or rational  value of n, provided there is a restriction on the value of x and y in the expansion (x +y)ⁿ

 Note the following characteristics of binomial expansion

1.                  In the expansion of (x + y)ⁿ, the number of term is (n + 1)

2.                  Every term take the form ⁿc_r x^{n– r}y^r

3.                  The degree of x and y are simultaneously on the decrease and increase respectively from left of both x and y equals n in each term

4.                  the coefficient of equidistant term from extremes are equal

ⁿc_r= ⁿc_{n – r}

e.g. (x + y)⁶ = x⁶ + 6x⁵y + 15x⁴ y² + 20x³y³ + 15x²y⁴ + 6xy⁵ + y⁶

The coefficient of x⁶ and y⁶ are equal, likewise the coefficient of x⁵y and xy⁵ are equal 

5.                  U_r = ⁿc_rx^n–ry^r denotes the general term (rth term) in the (r + 1) position of the expansion

U₁ = ⁿc₁x^n–1y = nx^{n – 1}y

U₂ = ⁿc₂x^n–2y² =  

U_n = ⁿc_nx^n–nyⁿ = yⁿ

Example

            Write down the binomial expansion of (x + 2y)⁵

Solution

            (x+2y)⁵=x⁵ + ⁵c₁(x)⁴(2y) + ⁵c₂(x)³(2y)² + ⁵c₃(x)²(2y)³ + ⁵c₄(x)(2y)⁴ + (2y)⁵  

Example

(a)                Write down the binomial expansion of

(b)               Use the expansion in (a) to evaluate (2.02)⁵ to 5s.f

Solution

Example

Expand in the power of  x, (1+3x + 3x²)³

Solution

             (1+ 3x +3x²)³ = [1+(3x+3x²)]³

               = 1+3(3x+ 3x² ) +3(3x +3x²)² +(3x +3x²)³

               = 1+9x +9x² +3(9x² +18x³+9x⁴)+(3x)³+(3x)²(3x²)+3(3x)(3x²)²+(3x²)³

               = 1+9x+9x²+27x²+54x³+27x⁴+27x³+27x⁴+81x⁵+27x⁶

               =  1+9x +36x²+81x³+54x⁴+81x⁵+27x⁶

   Example

 Find the term in x² and term in dependent of x in the expansion     

Solution

 

                                                    

 Now  x^6–r  = x^{– r + 6 – r}  = x^{6– 2r}

When this is x² then  6–2r = 2

                                r = 2 ( i.e. the 3^rd term )

Hence the term x² is  ⁶C₂ (3x)^{6– 2}  

                                         

If the term is to be independent of x then   6–2r =0

                                   therefore r = 3

(r + 1) th term is the  4th term in  the binomial  expansion 

             ⁶C₃ ( 3x) ^6–3     

             = 20(3x)³  which is independent of x

Example

Find the fifth term in the expansion of

Solution

The fifth term {i.e(r+1)th term which will be when r = 4} in the expansion (a + y)ⁿ is

 ⁿC₄(a^n–4)y⁴ =  a^n–4.y⁴

 Using n = 7, a = x²,  y = , the required term is

           

Example

      Find the coefficient of x⁸ in the expansion of (x²+3)¹²

Solution

      The (r + 1)^th  term in the expansion of (x²+3)¹² is    ¹²C_r (x²)^{12  –  r}(3)^r

                        =  ¹²C_r x^{24 – 2r} .(3)^r

If this be the required in x⁸, then

                        x^24–2r  = x⁸

                      24 – 2r = 8

                                    r = 8

the required coefficient is

                        ¹²C_r (3⁸) = . (3⁸)

                                       = 3,247,695

Example

Calculate without using tables, the value of

Solution

      

Using Pascal’s triangle for simplicity

On subtraction 1^st , 3^rd and 5^th term cancels out leaving 

           

Example

The expansion of  is used to find values (a) 0.995⁶  (b)   what value(s) of x should be substituted in each case

Example

  If the term of the expansion of (1+ px)ⁿ in ascending powers are 1+36x+540x²+ – – – + (px)ⁿ Find the value of p and n

Solution

(1+px)ⁿ = 1+36x +540x² – – – + (px)ⁿ

(1+px)ⁿ = 1+ⁿC₁ (Px) + ⁿC₂(px)² + – – – + (px)ⁿ        

             = 1+ n px +  p²x² + – – – + (px)ⁿ       

  1+36x+540x²  = 1+ n px +  p²x² + – – – + (px)ⁿ

Equating coefficient of like identities

            np = 36 ––––––––––– (1)

             = 540 ––––(2)

From equation (1)  –––(3)

Square both sides of equation (3),  substituted into equation (2)

Substitute 6 for n in equation (3)

                       

Hence n = 6,  p = 6

Example

Find the first four terms of the expansion    in ascending powers of x

Solution

      The expansion will be 1+ax+bx²+cx³ so we expand each binomial as far as x³ and  take the product of the terms

(1–3x)³ = 1+3(–3x)+3(–3x)+(–3x)³

              = 1–9x+27x²–27x³

 

Example

  The 2^nd, 3^rd and 4^th terms in the expansion of (a+b)ⁿ are 12, 60 and 160 respectively.  Find the value of a, b and n

Solution

(a+b)ⁿ = aⁿ + 12 + 60 + 160

Multiply (i) and (iii)

n²(n–1)(n–2)a^2n–4b⁴ = 11520 – – – – – (iv)

Square both side of eqn (ii)

           

Divide (iv) by (v)

           

Square (i)

            n²(a^n–1)²b² = 12²

            n² a^2n–2b² = 144 – – – – – – (vi)

Divide (ii) by (vi)

           

Substitute 6 for n

           

              a^–6 = 1 = 1^–6

                 a = 1

Substitute a = 1, n = 6 into (i)

                        na^n–1b = 12

                        6(1^6–1)b = 12

                                 6b = 12

                                  b = 2

                 a = 1, b = 2, n = 6

 

 

 

Example

     The first three terms in the expansion of a binomial are 32, 240, 720 respectively.  Find it

Solution

      Let the binomial be (a + b)ⁿ

The expansion of (a + b)ⁿ will be

(a+b)ⁿ = aⁿ+na^n–1b +  a^n–2b² + ----------+ bⁿ

 aⁿ = 32 ––––––––––(i)

   na^{n – 1}b = 240 –––––(ii)

    a^n–2b² = 720 ––(iii)

Multiply (i) and (iii) together

 (a^n–2)aⁿb² = 32×720

             a^2n–2b² = 23040

              n(n–1)a^2n–2b² = 46080 –––––––––––––(iv)

Square (ii)

            n²a^{(n – 1)2}b² = 240²

            n²a^2n–2b²  =  57600–––––––––––––––––––(v)

Divide (iv) by (v)

           

Substitute n = 5 into (i)

            a⁵ = 32 = 2⁵

            a = 2

Substitute a = 2, n  = 5 into (ii)

            na^{n – 1}b = 240

            5(2^5–1)b = 240

            5(2⁴)b = 240

                 80b = 240

                        b = 3

a  2, b = 3, n = 5

The required binomial expansion is, therefore (2+3)⁵

 

LINEAR APPROXIMATION

Consider the expansion

(1+x)ⁿ = 1+ⁿC₁ x + ⁿC₂ x³ + – – –    (higher powers of x)

           = 1+nx+ⁿC₂ x² + – – – –

if x is small then x², x³, – – – – will become smaller such that they become negligible

so (1+x)ⁿ  1+nx provided x is small enough to ignore x², x³, – – – – –

Similarly (1–x)ⁿ  1 – nx provided x is small

     This is called a LINEAR APPROXIMATION to the value of (1+x)ⁿ as we are only considering (1+nx), which is a linear function of x

Example

Find the linear approximation for

(a)        1.025⁶  (b)    

Solution

(a)        Using our formula

            (1+x)ⁿ  (1+nx)ⁿ

            1.025⁶ = (1+0.025)⁶  1+ 6(0.025)

                                             1+0.150   1.150

(b)       

                                            1+0.035  1.035

Example 

Find a linear approximation for (3.015)⁸

Solution

            Rewrite 3.015⁸ in the k(1+x)⁸

(3.015)⁸ = 3⁸  = 3⁸(1.005)⁸

                         = 3⁸(1+0.005)⁸

                        =3⁸(1+8(0.005))  3⁸(1+0.04)  6,823.44

                                                              

BINOMIAL EXPANSION FOR N<1

In the expansion (1–x)^–n where  n is positive all the terms will be positive as the (r+1)^th  term   will

           

 The following expansion would be obtained using the binomial theorem

This expansion is only valid if –1<x< 1 or 1 x 1 < 1

Example

Find the ascending powers of x of the following binomial expansion

(1)

 

 

 

 

Example

    Express f(x)  in partial fractions and hence give the first four terms in the expansion of f(x) in ascending power of x.  state the range of values of x for which the expansion is valid

Solution

     Let

Thus, we have

1 = A (4–x) + B(3–x)

Putting x = 3

            A = 1

Putting x = 4

            B = –1

Hence

Expanding

The range for which x in which the expansion is valid is /x/ <3 (–3<x<3)

 

Example

     Prove that, if x is small that terms in x³ and higher powers may be neglected, then

           

Solution

Example

     Find the possible values of f and h if the expansion in ascending powers of x up to the term x² of  is 1 – .  With these values of f and h, state the set of values of x for which the expansion is valid

Solution

 

EVALUATE WITH BINOMIAL EXPANSION

Use Binomial Expansion to evaluate

Examples

1.                  Show that if x is so small that x³ and higher power of x can be neglected then                           

2.                  Expand  in ascending powers of x as far as terms in x³

3.                  Obtain the expression of  in x as farm as the term in x³

Solution

1.                                                                               Expanding each terms       

                                                                                                                                                                                                                                                                    

2         

Alternative method to working the above example is shown