Calculating Coefficients without Pascal’s Triangle
Calculating Coefficients without Pascal’s Triangle
!!!
knknCk
n
Calculating Coefficients without Pascal’s Triangle
Proof
!!!
knknCk
n
Calculating Coefficients without Pascal’s Triangle
Proof
!!!
knknCk
n
Step 1: Prove true for k = 0
Calculating Coefficients without Pascal’s Triangle
10
CLHS n
Proof
!!!
knknCk
n
Step 1: Prove true for k = 0
Calculating Coefficients without Pascal’s Triangle
10
CLHS n
Proof
1!!
!!0!
nn
nnRHS
!!!
knknCk
n
Step 1: Prove true for k = 0
Calculating Coefficients without Pascal’s Triangle
10
CLHS n
RHSLHS
Proof
1!!
!!0!
nn
nnRHS
!!!
knknCk
n
Step 1: Prove true for k = 0
Calculating Coefficients without Pascal’s Triangle
10
CLHS n
RHSLHS
Proof
1!!
!!0!
nn
nnRHS
!!!
knknCk
n
Step 1: Prove true for k = 0
Hence the result is true for k = 0
0integer an is wherefor trueisresult theAssume:2 Step rrk
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
!1!1!.. 1
rnrnCei r
n
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
!1!1!.. 1
rnrnCei r
n
Proof
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
!1!1!.. 1
rnrnCei r
n
Proof11
1
rn
rn
rn CCC
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
!1!1!.. 1
rnrnCei r
n
Proof11
1
rn
rn
rn CCC
rn
rn
rn CCC
1
11
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
!1!1!.. 1
rnrnCei r
n
Proof11
1
rn
rn
rn CCC
rn
rn
rn CCC
1
11
!!
!!!1
!1rnr
nrnr
n
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
!1!1!.. 1
rnrnCei r
n
Proof11
1
rn
rn
rn CCC
rn
rn
rn CCC
1
11
!!
!!!1
!1rnr
nrnr
n
!!1
1!!1rnr
rnn
0integer an is wherefor trueisresult theAssume:2 Step rrk
!!!..
rnrnCei r
n
1for trueisresult theProve:3 Step rk
!1!1!.. 1
rnrnCei r
n
Proof11
1
rn
rn
rn CCC
rn
rn
rn CCC
1
11
!!
!!!1
!1rnr
nrnr
n
!!1
1!!1rnr
rnn
!!1
11!rnr
rnn
!!1
11!rnr
rnn
!!1
11!rnr
rnn
!!1
!rnr
rnn
!!1
11!rnr
rnn
!!1
!rnr
rnn
!1!1!
rnrn
!!1
11!rnr
rnn
!!1
!rnr
rnn
!1!1!
rnrn
Hence the result is true for k = r + 1 if it is also true for k = r
!!1
11!rnr
rnn
!!1
!rnr
rnn
!1!1!
rnrn
Hence the result is true for k = r + 1 if it is also true for k = r
Step 4: Hence the result is true for all positive integral values of n by induction
The Binomial Theorem
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
This extends to;
n
k
kknk
nn baCba0
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
This extends to;
n
k
kknk
nn baCba0
411Evaluate .. Cge
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
This extends to;
n
k
kknk
nn baCba0
411Evaluate .. Cge
!7!4!11
411 C
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
This extends to;
n
k
kknk
nn baCba0
411Evaluate .. Cge
!7!4!11
411 C
1234891011
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
This extends to;
n
k
kknk
nn baCba0
411Evaluate .. Cge
!7!4!11
411 C
1234891011
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
This extends to;
n
k
kknk
nn baCba0
411Evaluate .. Cge
!7!4!11
411 C
1234891011
3
The Binomial Theorem n
nnk
knnnnn xCxCxCxCCx 2
2101
n
k
kk
n xC0
integer positive a is and
!!!where n
knknCk
n
NOTE: there are (n + 1) terms
This extends to;
n
k
kknk
nn baCba0
411Evaluate .. Cge
!7!4!11
411 C
1234891011
330
3
(ii) Find the value of n so that;
(ii) Find the value of n so that;
85 a) CC nn
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
1135 ofexpansion in the 5th term theFind
baiii
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
1135 ofexpansion in the 5th term theFind
baiii
k
kkk b
aCT
35 1111
1
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
1135 ofexpansion in the 5th term theFind
baiii
k
kkk b
aCT
35 1111
1
4
74
115
35
baCT
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
1135 ofexpansion in the 5th term theFind
baiii
k
kkk b
aCT
35 1111
1
4
74
115
35
baCT
4
7474
11 35b
aC
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
1135 ofexpansion in the 5th term theFind
baiii
k
kkk b
aCT
35 1111
1
4
74
115
35
baCT
4
7474
11 35b
aC unsimplified
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
1135 ofexpansion in the 5th term theFind
baiii
k
kkk b
aCT
35 1111
1
4
74
115
35
baCT
4
7474
11 35b
aC unsimplified
4
78178125330b
a
(ii) Find the value of n so that;
85 a) CC nn
knn
kn CC
1358
nn
820
87 b) CCC nn
kn
kn
kn CCC
1
11
19n
1135 ofexpansion in the 5th term theFind
baiii
k
kkk b
aCT
35 1111
1
4
74
115
35
baCT
4
7474
11 35b
aC unsimplified
4
78178125330b
a
4
72088281250b
a
9
2
213in oft independen termObtain the
xxxiv
9
2
213in oft independen termObtain the
xxxiv
k
kkk x
xCT
213 929
1
9
2
213in oft independen termObtain the
xxxiv
k
kkk x
xCT
213 929
1
0 with termmeans oft independen term xx
9
2
213in oft independen termObtain the
xxxiv
k
kkk x
xCT
213 929
1
0 with termmeans oft independen term xx
0192 xxx kk
9
2
213in oft independen termObtain the
xxxiv
k
kkk x
xCT
213 929
1
0 with termmeans oft independen term xx
0192 xxx kk
60318
0218
kk
xxx kk
9
2
213in oft independen termObtain the
xxxiv
k
kkk x
xCT
213 929
1
0 with termmeans oft independen term xx
0192 xxx kk
60318
0218
kk
xxx kk
6
326
97 2
13
x xCT
9
2
213in oft independen termObtain the
xxxiv
k
kkk x
xCT
213 929
1
0 with termmeans oft independen term xx
0192 xxx kk
60318
0218
kk
xxx kk
6
326
97 2
13
x xCT
6
36
9
23C
9
2
213in oft independen termObtain the
xxxiv
k
kkk x
xCT
213 929
1
0 with termmeans oft independen term xx
0192 xxx kk
60318
0218
kk
xxx kk
6
326
97 2
13
x xCT
6
36
9
23C
Exercise 5D; 2, 3, 5, 7, 9, 10ac, 12ac, 13,
14, 15ac, 19, 25