Mathematics Dictionary
Dr. K. G. Shih
Binomial Expansion
Symbol Defintion
Example : Sqr(x) = square root of x
Q01 |
- Binomial Expansion : Coefficients C(n, r)
Q02 |
- Sequence from C(n+1, 2)
Q03 |
- Sequence from C(n+2, 3)
Q04 |
- Sequence from C(n+3, 4)
Q05 |
- References
Q01. Binomial Expansion : Coefficients C(n, r)
Deifintion of 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) + ....
It can be expressed as (x + y)^n = Sum[C(n,r)*(x^(n-r))*(y^r)]
Coefficients of expansion
C(n,r) = n*(n-1)*(n-2)*...*(n-r+1)/r!
Values
C(n,0) = Coefficient of first term = 1.
C(n,n) = Coefficient of last term = 1
C(n,1) = Coefficient of 2nd term = n
C(n,n-1) = Coefficient of last 2nd term = n
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Q02. Sequence from C(n+1, 2)
Questions
1. Write down the sequence
2. Find the nth term T(n)
3. Find the sum S(n)
Solutions
1. The sequence
n = 1 : C(2, 2) = 1
n = 2 : C(3, 2) = 3*2/(2!) = 3
n = 3 : C(4, 2) = 4*3/(2!) = 6
n = 4 : C(5, 2) = 5*4/(2!) = 10
n = 5 : C(6, 2) = 6*5/(2!) = 15
Etc.
This sequence is the triangular number patterns
It is 1, 3, 6, 10, 15, 21, .... as shown in Pascal triangle at r2 column
2. T(n)
T(n) = C(n + 1, 2) = n*(n + 1)/(2!)
3. S(n)
S(n) = Sum[C(n + 1), 2] = C(n + 2, 3)
S(n) = Sum[n*(n + 1)/2] = n*(n + 1)*(n + 2)/(3!)
Pscal triangle with r
Column ...... r0 r1 r2 r3 r4 r5 r6 r7 r8
---------------------------------
n=0 ......... 01
n=1 ......... 01 01
n=2 ......... 01 02 01
n=3 ......... 01 03 03 01
n=4 ......... 01 04 06 04 01
n=5 ......... 01 05 10 10 05 01
n=6 ......... 01 06 15 20 15 06 01
n=7 ......... 01 07 21 35 35 21 07 01
n=8 ......... 01 08 28 56 70 56 28 08 01
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Q03. Sequence from C(n + 2, 3) Questions
1. Write down the sequence
2. Find the nth term T(n)
3. Find the sum S(n)
Solutions
1. The sequence
n = 1 : C(3, 3) = 1
n = 2 : C(4, 3) = 4*3*2/(3!) = 4
n = 3 : C(5, 3) = 5*4*3/(3!) = 10
n = 4 : C(6, 3) = 6*5*4/(3!) = 20
n = 5 : C(7, 3) = 7*6*5/(3!) = 35
Etc.
This sequence is the triangular number patterns
It is 1, 4, 10, 20, 35, .... as shown in Pascal triangle at r3 column
2. T(n)
T(n) = C(n + 2, 3) = n*(n + 1)*(n + 2)/(3!)
3. S(n)
S(n) = Sum[C(n + 2), 3] = C(n + 3, 4)
S(n) = Sum[n*(n + 1)*(n + 2)/(3!)] = n*(n + 1)*(n + 2)*(n + 3)/(4!)
Pscal triangle with r
Column ...... r0 r1 r2 r3 r4 r5 r6 r7 r8
---------------------------------
n=0 ......... 01
n=1 ......... 01 01
n=2 ......... 01 02 01
n=3 ......... 01 03 03 01
n=4 ......... 01 04 06 04 01
n=5 ......... 01 05 10 10 05 01
n=6 ......... 01 06 15 20 15 06 01
n=7 ......... 01 07 21 35 35 21 07 01
n=8 ......... 01 08 28 56 70 56 28 08 01
Go to Begin
Q4. Sequence from C(n + 3, 4)
Questions
1. Write down the sequence
2. Find the nth term T(n)
3. Find the sum S(n)
Solutions
1. The sequence
n = 1 : C(4, 4) = 1
n = 2 : C(5, 4) = 5*4*3*2/(4!) = 5
n = 3 : C(6, 4) = 6*5*4*3/(4!) = 15
n = 4 : C(7, 4) = 7*6*5*4/(4!) = 35
n = 5 : C(8, 4) = 8*7*6*5/(4!) = 70
Etc.
This sequence is the triangular number patterns
It is 1, 5, 15, 35, 70, .... as shown in Pascal triangle at r4 column
2. T(n)
T(n) = C(n + 3, 4) = n*(n + 1)*(n + 2)*(n + 3)/(4!)
3. S(n)
S(n) = Sum[C(n + 3), 4] = C(n + 4, 5)
S(n) = Sum[n*(n + 1)*(n + 2)*(n + 3)/(4!)]
= n*(n + 1)*(n + 2)*(n + 3)*(n + 4)/(5!)
Pscal triangle with r
Column ...... r0 r1 r2 r3 r4 r5 r6 r7 r8
---------------------------------
n=0 ......... 01
n=1 ......... 01 01
n=2 ......... 01 02 01
n=3 ......... 01 03 03 01
n=4 ......... 01 04 06 04 01
n=5 ......... 01 05 10 10 05 01
n=6 ......... 01 06 15 20 15 06 01
n=7 ......... 01 07 21 35 35 21 07 01
n=8 ......... 01 08 28 56 70 56 28 08 01
Go to Begin
Q5. References
References :
Binomial theory
Definition
Binomial theory
in Analytic geometry Section 16
Psacal triangle
and sequence
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