Mathematics Dictionary

Mathematics Dictionary
Dr. K. G. Shih

Binomial Theorem and Pascal Triangle

Questions

AL 07 00 | - Outlines
AL 07 01 | - Defintion of Binomial Theorem
AL 07 02 | - Coefficients of Binomial expansaion
AL 07 03 | - Pascal triangle and coefficients of Binomial expansaion
AL 07 04 | - Sequence along r = 2 column : 1, 3, 06, 10, 15, ......
AL 07 05 | - Sequence along r = 3 column : 1, 4, 10, 20, 35, ......
AL 07 06 | - Sequence along r = 4 column : 1, 5, 15, 35, 70, .....
AL 07 07 | - Sequence along r = 5 column : 1, 6, 21, 56, 126 ....
AL 07 08 | - Find Sum[n^4] usimg Sum[C(n+3,4)] = C(n+4,5)
AL 07 09 | - Fibonacci's sequence in Pascal triangle
AL 07 10 | - Series from C(n,r)
AL 07 11 | - Constant oefficient in (x + 1/(x^2))^n
AL 07 12 | - Expand Sqr(1 + x^2) in series form
AL 07 13 | - Use Pascal triangle find coefficients of (x+y)^7
AL 07 14 | - Expand (x+1)^n
AL 07 15 | - Find coefficient (x^3)*(y^5) in expansion of (x+y)^n
AL 07 16 | - Expand 1/(1 + x^2)
AL 07 17 | - C(n,r-1) + C(n,r) = C(n+1,r)
AL 07 18 | - Coefficients of x^(r-1), x^r and x^(r+1) are 3 consecutive AP in (1+x)^n
AL 07 19 | - Coefficients of (x^p)*(y^q)*(z^r) in (x+y+z)^n
AL 07 20 | - Quiz and Answers

Answers

AL 07 01. Deifintion of Binomial Theorem

Factorial : n!

n! = 1*2*3*....*(n-1)*n
0! = 1
1! = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
5! = 1*2*3*4*5 = 120. It has one trailor zero.
Question : How many trailor zeros in 20!
- number of trailor zeros in n! is n/5 + n/(5^2) + n/(5^3) + ....
- The sum is only the integral part.
- For 20! : n = 20. Number of trialor zeros is 20/5 + 20/25 = 4
- For 25! : n = 25. Number of trialor zeros is 20/5 + 25/25 + 25/125 = 5

Binomial theorem

(x + y)^n = C(n,0)*x^n + C(n,1)*(x^(n-1))*(y)+ C(n,2)*(x^(n-2))*(y^2) + ....
Coefficients of expansion C(n,r) for r = 0,1,2,3....n
C(n,r) = n*(n-1)*(n-2)*...*(n-r+1)/r! = coefficient of (r+1)th term
- C(n,0) = C(n,n) = 1 .................. coeficient of 1st term
- C(n,1) = n ........................... coeficient of 2nd term
- C(n,2) = n*(n-1)/2! .................. coeficient of 3nd term
- ...
- C(n,n-2) = C(n,2) = n*(n-1)/2! ....... coeficient of 3rd last term
- C(n,n-1) = C(n,1) = n ................ coeficient of 2nd last term
- C(n,n-0) = C(n,0) = 1 ................ coeficient of last term
It can be expressed as (x+y)^n = Sum[C(n,r)*(x^(n-r))*(y^r)]
- Power of x : n, (n-1), (n-2), ... 3, 2, 1, 0
- Power of y : 0, 1, 2, 3, ........ (n-2), (n-1), n
- Sum of power x and power y is n.
It has (n + 1) terms.

Properties of C(n,r)

Sum of the coeefficients = C(n,0) + C(n,1) + .... = 2^n.
- Let x = y = 1, we have
- C(n,0) + C(n,1) + .... = 2^n.
Sum of the coeefficients of odd terms = sum of coefficients of even terms.
- Let x = 1 and y = -1, we have
- C(n,0) + C(n,2) + ... = C(n,1) + C(n,3) + ...
C(n,r) = C(n,n-r)
- LHS = n*(n-1)*...*(n-r+1)/r!
- LHS = n*(n-1)*(n-2)*...*(n-r+1)*(n-r)!/((r!)*(n-r)!)
- LHS = n!/((r!)*(n-1)!)
- RHS = n*(n-1)*(n-2)*...*(n-(n-r)+1)/(n-1)!
- RHS = n*(n-1)*(n-2)*....*(r+1)/(n-1)!
- RHS = n*(n-1)*(n-2)*....*(r+1)*r!/((n-1)!)*r!)
- RHS = n!/((r!)*(n-1)!)
- Since LHS = RHS
- Hence C(n,r) = C(n,n-r).

Computer program : Find n!, P(n,r) and C(n,r)

Probability Program 01 01 : P(n,n) = n!
Program 01 02 : C(n,r)

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AL 07 02. Coefficients of Binomial expansion

Expand (x+y)^2

(x+y)^2 = x^2 + 2*x*y + y^2
Coefficients : 1, 2, 1

Expand (x+y)^3

(x+y)^3 = x^3 + 3*(x^2)*y + 3*x*y^2 + y^3
Coefficients : 1, 3, 3, 1

Expand (x+y)^4

(x+y)^4 = x^4 + 4*(x^3)*y + 6*(x^2)*(y^2) + 4*x*y^3 + y^4
Coefficients : 1, 4, 6, 4, 1

Example : Find coefficients in (x + y)^4 from coefficients in expansion of (x+y)^3

Coefficients in (x+y)^3 : C(3,0)=1, C(3,1)=3, C(3,2)=3, C(3,3)=1
Coefficients in (x+y)^4 : 1, 4, 6, 4, 1
Coefficients of 1st term in (x+Y)^4 = C(4,0) = C(3,0) = ................ 1
Coefficients of 2nd term in (x+Y)^4 = C(4,1) = C(3,0) + C(3,1) = 1 + 3 = 4
Coefficients of 3rd term in (x+Y)^4 = C(4,2) = C(3,1) + C(3,2) = 3 + 3 = 6
Coefficients of 4th term in (x+Y)^4 = C(4,3) = C(3,2) + C(3,3) = 3 + 1 = 4
Coefficients of 4th term in (x+Y)^4 = C(4,4) = C(3,3) + C(3,0) = 3 + 1 = 1

Example : Find coefficients in (x+y)^5 and in (x+y)^6

Coeff in (x+y)^4 : 1, 04, 06, 04, 01
Coeff in (x+y)^5 : 1, 05, 10, 10, 05, 1
Coeff in (x+y)^6 : 1, 06, 15, 20, 15, 6, 1

Example : Find coefficient of (x^5)*(x*4) in expansion of (x+y)^n.

Find C(n,r)
- C(n,r) is the coefficient (x^(n-r))*(y^r) in expansion of (x+y)^n
- x power + y power = n.
Find n and r
- n = 5 + 4 and r = 4.
- Coeficient of (x^5)*(y^4) = C(9,4)
- C(9,4) = 9*8*7*(9-4+1)/4!
- C(9,4) = (9*8*7*6)/(4*3*2*1)
- C(9,4) = 126.

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AL 07 03. Coefficients of binomial expansion and Pascal triangle

[Defintion] Pascal triangle : It is coeff of binomial expansion

1 .................0..... Coeff in (x+y)^0
1, 01 ................... Coeff in (x+y)^1
1, 02, 01 ............... coeff in (x+y)^2
1, 03, 03, 1 ............ coeff in (x+y)^3
1, 04, 06, 04, 1 ........ coeff in (x+y)^4
1, 05, 10, 10, 5, 1 ..... coeff in (x+y)^5

[Pscal triangle]

[Example] Find coefficient of (x^3)*(y^5) in expansion of (x+y)^n

The power n = 3 + 5 = 8
For y^5 we know r = 5
At row n = 8 and r = 5 we have C(n,r) = C(8,5) = 70

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AL 07 04. Sequence in Pascal triangle

Sequence along r0

r0 sequence : 1, 1, 1, 1, ....
nth term :T(n) = 1
Sum of n terms = S(n) = Sum[n] = n

Sequence along r1

r0 sequence : 1, 2, 3, 4, 5, ....
nth term :T(n) = n
Sum of n terms = S(n) = Sum[n] = n*(n+1)/2

Sequence along r2

r2 sequence : 1, 3, 6, 10, 15, 21, 28, 36, ....

T(n) = nth term = n*(n+1)/2

S(n) = Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6

Sum[C(n+1, 2)] = C(n+2, 3)

This is known as numbers in triangular number pattern

Properties of this pattern
- Properties of sequence of numbers in triangular sequence

[Example] Prove that Sum[n*(n+1)/2] = n*(n+1)*(n+2)/3!

Sum[n*(n+1)/2]
= Sum[n^2/2]+Sum[n/2]
= (Sum(n^2) + Sum[n])/2
= (n*(n+1)*(2*n+1)/6 + n*(n+1)/2)/2
= n*(n+1)*(2*n+1)/12 + n*(n+1)/4
= n*(n+1)*(2*n+1)/3+1)/4
= n*(n+1)*(2*n+4)/12
= n*(n+1)*(n+2)/3! where 3!=1*2*3=6

[Example] Prove that Sum[C(n+1,2)] = C(n+2,3)

Since C(n+1,2) = n*(n+1)/2!
Since C(n+2,3) = n*(n+1)*(n+2)/3!
Hence Sum[C(n+1,2)] = C(n+2,3)

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AL 07 05. What is r3 series ?

Sequence

Squence along r3 column : 1, 4, 10, 20, 35, 56, ....
T(n) = nth term = n*(n+1)*(n+2)/3!
S(n) = Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
Hence Sum[C(n+2, 3)] = C(n+3, 4)
- 1st difference is r2 series
- 2nd difference is r1 series
- 3nd difference is common difference d=1

[Example] Prove that Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!

Sum[n*(n+1)*(n+2)/6]
= Sum[n^3/6]+Sum[3*n^2/6]+Sum[2*n/6]
= (n*(n+1)/2)/6 + n*(n+1)*(2*n+1)/12 + n*(n+1)/12
= n*(n+1)*[(n*(n+1)/24 + (2*n+1)/12 +1/12]
= n*(n+1)*(n^2+5*n+6)/24
= n*(n+1)*(n+2)*(n+3)/4! where 4!=1*2*3*4
Hence Sum[C(n+2,3)] = C(n+3,4)

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AL 07 06. What is the sequence along r4 column ?

Sequences

Sequence along r4 column : 1, 5, 15, 35, 70, ....
T(n) = nth term = n*(n+1)*(n+2)*(n+3)/4!
S(n) = Sum[T(n)] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
Hence Sum[C(n+3, 4)] = C(n+4, 5)
Difference
- 1st difference is r3 series
- 2nd difference is r2 series
- 3rd difference is r1 series
- 4th difference is common difference d=1

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AL 07 07. What is the sequence along r5 column ?

[Definition]

Sequence along r5 column
T(n) = nth term = n*(n+1)*(n+2)*(n+3)(n+4)/5!
S(n) = Sum[T(n)] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
Hence Sum[C(n+4, 5)] = C(n+5, 6)
Difference
- 1st difference is r4 series
- 2nd difference is r3 series
- 3rd difference is r2 series
- 4th difference is r1 series
- 5th difference is common difference d=1

[Coclusion]

Sum[C(n+0,1)] = C(n+1,2)
Sum[C(n+1,2)] = C(n+2,3)
Sum[C(n+2,3)] = C(n+3,4)
Sum[C(n+3,4)] = C(n+4,5)
Sum[C(n+4,5)] = C(n+5,6)

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AL 07 08. Find Sum[n^4] by using Sum[C(n+3,4)] = C(n+4,5)

Method

Sum[C(n+3),4] = Sum[n*(n+1)*(n+2)*(n+3)/4!] = C(n+4,5)
(Sum[n^4] + 6*Sum[n^3] + 11*Sum[n^2] + 6*Sum[n])/24 = C(n+4,5)
We know that
- Sum[1] = n.
- Sum[n] = n*(n+1)/2.
- Sum[n^2] = n*(n+1)*(2*n+1)/6.
- Sum[n^3] = (n*(n+1)/2)^2
The answer is Sum[n^4] = (6*n^5 + 5*n^4 + 10*n^3 - n)/30

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AL 07 09. Fibonacci's sequence in Pascal triangle
Text and diagram

in Fibonacc's pattern

Example

Fibonacci's Sequence : 1, 1, 2, 3, 5, 8, 13, ...
Recursion formula :
- T(0) = 0.
- T(1) = 1.
- T(n+2) = T(n+1) + T(n)

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AL 07 10. Series from C(n,r)

[Sum in C(n,r) Form]

Sum[C(n+1,2)] = C(n+2,3)
Sum[C(n+2,3)] = C(n+3,4)
Sum[C(n+3,4)] = C(n+4,5)
Sum[C(n+4,5)] = C(n+5,6)

[Sum in factor Form]

Sum[1] = n
Sum[n] = n*(n+1)/2!
Sum[n*(n+1)/2!] = n*(n+1)*(n+2)/3!
Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
Sum[n*(n+1)*(n+2)*(n+3)/4!] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!

[Other series]

Study subject | Sum[n^2] = n*(n+1)*(2*n+1)/6
Study subject | Sum[n^3] = (n*(n+1)/2)^2
Study subject | Binomial theorem in calculus
Sum[n^4] = (6*n^5 + 5*n^4 + 10*n^3 - n)/30

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AL 07 11. Constant oefficient in (x + 1/(x^2))^n

The rth term

rth term = C(n,r)*(x^(n-r))*(1/(x^2))^r)
= C(n,r)*(x^(n-3*r))

Constant term

Power n - 3*r = 0 and r is integer
Hence r = n/3
Hence n must be multiple of 3.

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AL 07 012. Expand Sqr(1+x^2) into series

By binomial theory

Sqr(1 + x^2)
= (1 + x^2)^(1/2)
= 1 +(1/2)*(x^2) +((1/2)*(1/2-1)/2!)*((x^2)^2) +((1/2)*(1-1/2)*(1/2-2)/3!)*(x^2)^3 + ...
= 1 + (1/2)*(x^2) - (1/((2^2)*2!)*(x^4) + (1*3/(2^3)*3!)*(x^6) + ...

Find Sqr(2) using above series

Let x = 1
Sqr(2) = 1 + 1/2 - 1/8 + 1/16 - (1*3*5)/((2^4)*4!) + (1*3*5*7)/((2^5)*5!) - ...
= 1 + 0.5 - 0.125 + 0.0625 - 0.03906 + 0.02734 - 0.025078....
= 1.40070
Error = 1.41421 - 1.40070 = 0.01351 =0.95%
What is the error if we use the 8th term ?

Find Sqr(26) to 4th term by binomail expansion

Sqr(26) = Sqr(25 + 1)
= 5*(1 + (1/5)^2)
= 5*(1 + (1/2)*(1/5)^2 - (1/8)*(1/5)^4 + (1/16)*(1/5)^6
= 5*(1 + 0.02 - 0.0002 + 0.000028) = 5*(1.020228) = 5.099114

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AL 07 13. Use Pascal triangle write down coefficient of expansion terms of (x+y)^7

Study subject | Solution see AL 07a 13

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AL 07 14. Expansion of (x+1)^n

Expansions of (x+1)^n

(x+1)^1 = x^1 + 1*x^0
(x+1)^2 = x^2 + 2*x^1 + 1.
(x+1)^3 = x^3 + 3*x^2 + 3*x + 1.
(x+1)^4 = x^4 + 4*x^3 + 6*x^2 + 4*x + 1.
(x+1)^5 = x^5 + 5*x^4 + 10*x^3 + 10*x^2 + 5*x + 1

Notes

x^0 = 1.
x^1 = x.

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AL 07 15. Find coefficient 0f (x^3)*(y^5) in expansion (x+y)^n

[Defintion] Pascal triangle : It is coeff of binomial expansion

1, 2, 1 are ceeficients of expansion of (x+y)^2
1, 3, 3, 1 are ceeficients of expansion of (x+y)^3
1, 4, 6, 4, 1 are ceeficients of expansion of (x+y)^4

[Pscal triangle]

[Example] Find coefficient of (x^3)*(y^5) in expansion of (x+y)^n

The x power and y power = n = 3 + 5 = 8.
For y^5 we know it is in (5+1)th term (r =5).
At row n=8 and r=5 we have C(n,r) = C(8,5) = 56.
Hence the coefficient of (x^3)*(y^5) is 56.
Or use formula : C(8,5) = C(8,3) = (8*7*6)/(3*2*1) = 56.

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AL 07 16. Expansion of 1/(1 + x^2)

Expand 1/(1+x^2) into series

1/(1+x^2)
= (1 + x^2)^(-1)
= 1 +(-1)*(x^2) +((-1)*(-1-1)/(2!))*((x^2)^2) +((-1)*(-1-1)*(-1-2)/(3!))*((x^2)^3)+ ...
= 1 - x^2 + x^4 - x^6 + x^8 + .....

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AL 07 17. C(n,r-1) + C(n,r) = C(n+1,r)

For n = 1, 2, 3 : Proof using Pascal triangle

n = 1 : C(1,0) = 1, C(1,1) = 1
n = 2 : C(2,0) = 1, C(2,1) = 2, C(2,2) = 1
n = 3 : C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C(3,3) = 1
n = 4 : C(4,0) = 1, C(4,1) = 4, C(4,2) = 6, C(4,3) = 4, C(4,4)
For n = 1
- C(1,0) + C(1,1) = 2 and C(2,1) = 2. Hence C(1,0) + C(1,1) = C(2,1)
For n = 2
- C(2,0) + C(2,1) = 3 and C(3,1) = 3. Hence C(2,0) + C(2,1) = C(3,1)
- C(2,1) + C(2,2) = 3 and C(3,2) = 3. Hence C(2,1) + C(2,2) = C(3,2)
For n = 3
- C(3,0) + C(3,1) = 4 and C(4,1) = 4. Hence C(3,0) + C(3,1) = C(4,1)
- C(3,1) + C(3,2) = 6 and C(4,2) = 6. Hence C(3,1) + C(3,2) = C(4,2)
- C(3,2) + C(3,3) = 4 and C(4,3) = 4. Hence C(3,2) + C(3,3) = C(4,3)

For r = r+1 : We expect C(n,r) + C(n,r+1) = C(n+1,r+1)

Study subject | Proof : See AL 07a 01

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AL 07 18. Coefficients of x^(r-1), x^r and x^(r+1) are 3 consecutive AP in (1+x)^n

Question : Prove that n^2 - n*(4*r + 1) +4*r^2 - 2 = 0

Coeff of x^(r - 1) = a = (n)*(n-1)*.....*(n-r+2)/(r-1)!
Coeff of x^(r - 0) = b = (n)*(n-1)*.....*(n-r+2)*(n-r+1)/(r-0)!
Coeff of x^(r + 1) = c = (n)*(n-1)*.....*(n-r+2)*(n-r+1)*(n-r+0)/(r+1)!
Coeff a,b,c are consecutive terms in AP, hence b = (a + c)/2
b = (n-r+1)/r! = (1/(r-1)! + (n-r+1)*(n-r)/(r+1)!)/2
Since (r+1)! = ((r-1)!)*r*(r+1) and r! = ((r-1)!)*r
(n-r+1)/r = ( 1 + (n-r+1)*(n-r)/(r*(r+1))
Hence (n-r+1)*r*(r+1) = r*(r+1) + (n-r+1)*(n-r)
Simplify we have n^2 - n*(4*r + 1) +4*r^2 - 2 = 0

Find a, b, c if n = 14

14^2 - 14*(4*r + 1) + 4*r^2 - 2 = 0
196 - 56*r - 14 + 4*r^2 - 2 = 0
4*r^2 - 56*r + 180 = 0
r^2 - 14 + 45 = 0
(r - 5)*(r - 9) = 0
When r = 5
- a = C(14,4) = 14*13*12*11/4!
- b = C(14,5) = 14*13*12*11*10/5!
- c = C(14,6) = 14*13*12*11*10*9/6!
- Verify : b = 14*13*12*11*10/(4!)*5 = 2*(14*13*12*11)/4!
- a + c = (14*13*12*11/(4!))*(1 + 10*9/(5*6)) = 4*(14*13*12*11/(4!))
- b = (a + c)/2
When r = 9 : we use C(n,r) = C(n,n-r)
- a = C(14, 9-1) = C(14, 14-08) = C(14,6)
- b = C(14, 9-0) = C(14, 14-09) = C(14,5)
- c = C(14, 9+1) = C(14, 14-10) = C(14,4)
- The results same as when r = 5 : b unchange and a, c are interchanged