Pascal's Triangle coefficients

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
21st September 2019
1.)

Pascal’s Triangle coefficients
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Pascal’s Identity has it that
( n
k ) =
( n−1
k−1 )+ ( n−1
k )
Taking row six, in the position 2 of row 6 we have (6
2 )=15.Thus with n=6 we have
∑
k=0
6
(6
k )= (6
0 )+ (6
1 )+ (6
2 )+ (6
3 )+
(6
4 )+
(6
5 )+ (6
6 )
But ( 6
0 ) =1∧( 6
6 )=1
(6
1 )=
(5
0 )+ (5
1 )=1+5=6
( 6
2 ) =
( 5
1 ) +
( 5
2 )=5+10=15
(6
3 )=(5
2 )+(5
3 )=10+10=20
( 6
4 )= ( 5
3 )+ ( 5
4 )=10+5=15
(6
5 )=(5
4 )+(5
5 )=5+1=6

∑
k=0
6
(6
k )= (6
0 )+ (6
1 )+ (6
2 )+ (6
3 )+(6
4 )+(6
5 )+ (6
6 )
∑
k=0
6
( 6
k )=1+6+15+20+15+6+1=64
∑
k=0
6
(6
k )=64
Taking row five, in the position 3 of row 6 we have ( 5
3 )=10.Thus with n=5 we have
∑
k=0
5
( 5
k )= ( 5
0 ) +
( 5
1) + ( 5
2 ) +
( 5
3 ) + ( 5
4 ) + ( 5
5 )
But (5
0 )=1∧(5
5 )=1
( 5
1 )= ( 4
0 )+ ( 4
1 )=1+4=5
(5
2 )= (4
1 )+ (4
2 )=4+6=10
( 5
3 )= ( 4
2 )+ ( 4
3 )=6+4=10
(5
4 )= (4
3 )+ (4
4 )=4+1=5
∑
k=0
5
( 5
k )= ( 5
0 ) +( 5
1) + ( 5
2 ) +( 5
3 ) + ( 5
4 ) + ( 5
5 )
∑
k=0
5
( 5
k )=1+5+10+10+5+1=32
∑
k=0
5
(5
k )=32
∑
k=0
6
(6
k )=64 And ∑
k=0
5
(5
k )=32
64=2×32