Binomial Coefficients Proofs

Notation Issues? Practise moving between representations:  n  n  n  n  n (1 + x)n =      x    x 2 ...  x r ...  x n  0  1   2 r  n = n C0  nC1 x  nC2 x 2 ... nCr x r ... nCn x n n

=

 n

  r  x

r

r 0 n

=

n!

 r!(n  r )! x

r

r 0 n

=

 r 0

n

Cr x r

Students will find proofs of general results easier if they change the question immediately to the notation with which they feel most comfortable.

Hints, Tips and Tricks: 1. If the result involves a sum of nCr ’s with no x’s then a value has probably been substituted in for x. 2. If the result involves a sum of nCr ’s with alternating signs, begin by considering  n  n  n  n  n (1  x)n =      x    x 2 ...( 1) r   x r ...( 1) n   x n and substitute for x.  n  0  1   2 r 3. If the result involves powers of a number (say a) as well as nCr ’s, the value of x was probably a. 4. If the nCr ’s are being multiplied by their corresponding r value, begin by differentiating both sides  n  n  n  n  n of (1 + x)n =      x    x 2 ...  x r ...  x n .  2 r  n  0  1  5. If the nCr ’s are being divided by their corresponding (r + 1) value, begin by integrating both sides of  n  n  n  n  n (1 + x)n =      x    x 2 ...  x r ...  x n , and remember the constant of integration.  n  0  1   2 r Exercise:

 n  n  n  n  n 1. Prove that         ...  ...   2 n .  0  1   2 r  n  n  n  n  n  n 2. Prove that            ... ( 1) n    0 .  n  0  1   2  3 n

3. Prove that

 n

 r  r   n  2

n 1

.

r 0

1  n 2 n 1  1 .    n 1 r 0 r  1  r  n

Prove that

Considering the expansion of 1  x 

n

(i)

n n n What is       ...    ? n  0  1 

(ii)

 n n n n What is    2    4   ...  2n   ? n 0 1  2

n n n  n  n n n n (iii) If n is even, why is         ...             ...    ?  n  1   3   5  0  2 4  n 1

Now try the Supplementary Problems on page 4

Supplementary Work

5. Hint: Use the identity

 2n   2n     n  r  1   nr  n  r  1

n  r

Identities with Binomial Coefficients n  n       ,  k n k    

(Part 2)

for 0  k  n

(1)

Prove

(2)

By considering the identity  x  1  n  1  n   n        k   k   k 1      

(3)

(4)

n

 1 1   x 

(i)

 n  n   n  n      ...         2  0   3  1 

(ii)

 n   0

n    1

2

  x  1 x  1 , prove n

for 1  k  n.

By considering 1 x 

2

n 1

n     2

2

n

express in simplest form

 1

n  n

 n  ...     n

n       n n  2

2

if n is even.



By equating coefficients on both sides of 1  x  1  x   1  x 2 n

n

Show

  1

k 0

k

n   k 

2

 0 if n is odd

n



n

Identities with Binomial Coefficients Solutions (Part 2) n  n       , for 0  k  n  k n k     n  n  LHS    RHS   k   n  k      n! n! n!     n   n  k   ! n  k  !  n  k  !k !  n  k !k !

(1)

Prove

(2)

By considering the identity  x  1

n 1

  x  1 x  1 , prove n

 n  1  n   n       for 1  k  n.  k   k   k 1        n  1 n 1 k LHS    = coefficient of x in the expansion of 1  x   k  Given that 1  x 1  x   1  x   x 1  x  then the coefficient of x k of the RHS is the sum n

n

n

of the coefficient of x k in 1  x  and the coefficient of x k 1 in 1  x  n

n  n      k   k  1  n  1  n   n        k   k   k  1 i.e.

(3)

By considering 1 x 

n

 1 1   x 

n

express in simplest form

n n n x 2  1   1  1  x   x  1  1  x  1    xn xn  x n n k n k n 2 nk  1 x    1   x 2n2k       k 0 k 0 k  k    n x xn n k n    1   x n 2 k k 0 k  n

n

n

(i)

 n  n   n  n      ...     0   3  2      1 

 1

n  n

n       n n  2

n n n 2   is the coefficient of x in 1  x  and   is the constant in 2 0

 1 1    x

n

 n n 1 n      is the coefficient of a term in x 2 in 1  x  1   .  x  20  n

n n n  1 3 1   is the coefficient of x in 1  x  and    is the coefficient of x in 1    x 3  1 

n

 n n 1 n       is the coefficient of a term in x 2 in 1  x  1   .  x  3  1  n

 n  n   n  n  1 n  n  n  n 2 So         ...   1    is the coefficient of x in 1  x  1    x  2  0   3  1   n  n  2  n

Now what is the coefficient of x 2 in

  1 k 0

k

n

 n  n2k ?  x k 

n  2k  2  2k  n  2 k 

n2 2

 n  n2  n n  n n n  n n    So if n is even then           ...   1       1 2  n  2   n n 2 0 3 1 2           2   n  n   n  n  n n n  BUT if n is odd then there is no x 2 term and so         ...   1    0  2  0   3 1   n n  2

0   n n  n n n  n n   n  n2             ...   1          2   0   3  1   n   n  2   1 2  n  2    2 

n odd n even

(ii)

 n   0

2

n    1

2

n     2

2

 n  ...     n

2

if n is even.

n n n   is the constant term in 1  x  and   is the constant in 0 0 2

 1 1    x

n

n 1 n    is a constant term in 1  x  1   .  x 0 n

n n n  1 1   is the coefficient of x in 1  x  and    is the coefficient of x in 1    x 1  1  2

n 1 n     is a constant term in 1  x  1   .  x 1 

2

2

n

2

n n n 1 n So       ...    is the constant term in 1  x  1   .  x  0  1  n n

n

Now what is the constant term in

  1 k 0

n  2k  0  2k  n n  k     n even  2 2 2 2 n n n n n 2        ...      1  n     0  1  n 2

k

 n  n2k ?  x k 

n

(4)



By equating coefficients on both sides of 1  x  1  x   1  x 2 n

n

  1

Show

k 0

1  x  1  x  n

n

k

n   k 



n

2

 0 if n is odd

 n n j   n  k n     x     1   x k  k    j 0  j    k 0

1  x     1 2 n

n

n

r 0

2

r

 n  2r  x r 

2

2

2

n n n n n Now   1          ...   1   k 0  k   0  1  n  n  n   n  n  n  n  n           ...   1     n  0   n  11   0  n  n

k

n n n n n   is the coefficient of x in 1  x  and   is the is the constant term in 1  x  0 n 2

n n n    is the coefficient of a term in x n in 1  x  1  x  . 0

 n  n n n n 1   is the coefficient of x in 1  x  and    is the coefficient of x in 1  x  .  n  1 1  2

n n n     is the coefficient of a term in x n in 1  x  1  x  . 1 

2

2

2

n n n n n n        ...   1   is the coefficient of the term in x n in 1  x  1  x  .  0  1  n

n n r n Now, what is the coefficient of the term in x n in the expansion of 1  x 2     1   x 2 r ? r 0 r 

0

Ans:

It was given that n is odd and all the terms have even powers, there are no terms with odd powers. 2

2

2

n n n n        ...   1    0  0  1  n