show that the expression 2n!/2^n(n!) is an integer for all n greater than or equ
ID: 1720037 • Letter: S
Question
show that the expression 2n!/2^n(n!) is an integer for all n greater than or equal to (b) Use part (a) to obtain the inequality 2" (n!) s (2h)! for al n l 9. Establish the Bernoulli inequality: If 1 +a >0, then for all n 1. 10. For all n 2 1, prove the following by mathematical induction: 12 22 32 2n 11. Show that the expression (2n)!/2"n! is an integer for all n 2 0. 12. Consider the function defined by 3n 1 for n odd T(n)=| 2- for n even The 3n +1 conjecture is the claim that starting from any integer of iterates T (n), T (T(n)), T (T (T(n))),..., eventually reaches the quently runs through the values 1 and 2. This has been verified foraExplanation / Answer
show that the expression 2n!/2^n(n!) is an integer for all n greater than or equal to 0.
Proof by induction on n;
n =1 . the number is 2/2 =1, so verified.
Assume the statement true for n.
The corresponding number for (n+1) is
(2(n+1))!/[ 2(n+1) (n+1)!]
= [(2n+1) (2n+2)/2(n+1)] [(2n!) / 2n (n!)]
= Integer x Integer (the second follows from induction hypothesis)
= Integer.
So we are done
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