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 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 nr 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 nk 1 x 1 x 2n2k 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 20 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 n2k ? x k
n 2k 2 2k n 2 k
n2 2
n n2 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
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