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HI
I just recieved some problems today that I have to complete, but once I got home I noticed that I couldn't answer the questions, I can answer a couple of the questions but for the most part I am lost.
Thank You
1. a. Given t1=1, list the next five terms for the recursion formula tn= n*tn-1
b. In the sequence, tk, is a factorial number, often written as k! Show that
Tk=k!
=k(k-1)(k-2)......(2)(1).
c. Explain what 8! means. Evaluate 8!
d. Explain why factorial numbers can be considered an iterative process.
e. Note that
(2to the power of5)(5!)
= (2*2*2*2*2)(5*4*3*2*1)
= (2*5)(2*4)(2*3)(2*2)(2*1)
=10*8*6*4*2
which is the product of the first five even positive integers. Write a formula for the product of the first n even positive integers. Explain why your formula is correct.
f. Write 10!/(2to the power of 5)(5!) as a product of consecutive odd integers.
g. Write a factorial formula for the product of
i) the first six odd positive integers
ii) the first ten odd positive integers
iii) the first n odd positive integers
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the first question: t2=2*1, t3=3*2*1=6,t4=4*3*2*1=24, t5=5*4*3*2*1=120, t6=6*5*4*3*2*1=720.
b. you can prove it by inductive method! when k=1, tk=1( it's correct)
suppose that clause is right when k=a or ta=a!(a>1),we have to prove that is also right when K=a+1 or ta+1=(a+1)!
because ta=a! ,so that ta=a(a-1)(a-2)(a-3)......3.2.1 (1)
but ta+1= (a+1).ta=(a+1)a(a-1)(a-2)......3.2.1=(a+1)! the clause is proved) . we conclude: tk=k! with k is belong to N* .
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i had a mistake, please replace the image by "="
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