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Assignment 3
Question 1
Check the case when n = 5. The conjecture states:
2 n
n
2 n − 2
Show that:
8
Calculating the values:
10 × 9 × 8 × 7 × 6
5 × 4 × 3 × 2 × 1
Hence n = 5 checks out because
8
Assuming the conjecture holds for some positive integer k ≥ 5, i.e.,
2 k
k
2 k − 2
Prove that the conjecture also holds for k+1, i.e.,
2 ( k + 1 )
k + 1
2 ( k + 1 )− 2
Calculating the binomial coefficient on the left side:
2 ( k + 1 )
k + 1
( 2 k + 2 )!
( k + 1 )! ( 2 k + 2 −
k + 1
( 2 k + 2 )!
( k + 1 )!
k + 1
Using the inductive assumption:
2 k
k
2 k − 2
This means that:
( 2 k )!
k! k!
2 k − 2
Rewriting the expression for
2 ( k + 1 )
k + 1
using this information:
2 ( k + 1 )
k + 1
( 2 k + 2 )!
( k + 1 )!
k + 1
2 ( k + 1 )
k + 1
( 2 k + 2 )( 2 k + 1 )( 2 k )!
( k + 1 )!
k + 1
Divide both sides by
2 k
k
2 ( k + 1 )
k + 1
2 k
k
( 2 k + 2 )( 2 k + 1 )( 2 k )!
( k + 1 )! ( k + 1 )!
( 2 k )!
k !k!
Simplify:
2 ( k + 1 )
k + 1
2 k
k
( 2 k + 2 )( 2 k + 1 )
( k + 1 )( k + 1 )
Using the inductive assumption:
( 2 k + 2 )( 2 k + 1 )
( k + 1 )( k + 1 )
2 k − 2
( 2 k + 2 )( 2 k + 1 )
( k + 1 ) ( k + 1 )
( 2 k + 2 )( 2 k + 1 )
( k + 1 )
k + 1
2 k − 2
∙ ( 2 k + 2 )
RHS of the inequality for k+1:
2 ( k + 1 )− 2
2 k
2
2 k
See that, 4 ∙ 2
2 k
2 k − 2
∙ ( 2 k + 2 ) since 4 > 2k + 2, and therefore:
2 k
2 k − 2
∙ ( 2 k + 2 )
Combining both inequalities:
( 2 k + 2 )( 2 k + 1 )
( k + 1 )
k + 1
2 ( k + 1 )− 2
By mathematical induction, the conjecture is proven for all natural numbers n ≥ 5.
Question 2
Check the case when n = 2. The conjecture states:
n <
n
∙n
