SOLUTION: 4 A recurrance relationship is such that Un+1=a/4Un+12.
4(a).If U0=16 show clearly that U2=aČ+3a+12
4(b).Hence find a if U2=30 and a>0
Can you provide full working with solu
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-> SOLUTION: 4 A recurrance relationship is such that Un+1=a/4Un+12.
4(a).If U0=16 show clearly that U2=aČ+3a+12
4(b).Hence find a if U2=30 and a>0
Can you provide full working with solu
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Question 388620: 4 A recurrance relationship is such that Un+1=a/4Un+12.
4(a).If U0=16 show clearly that U2=aČ+3a+12
4(b).Hence find a if U2=30 and a>0
Can you provide full working with solution.
Many Thanks,
Andy Found 2 solutions by jsmallt9, robertb:Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website! Un+1 =
a. If show clearly that .
Using and the equation above we can find find and then use that to find . We start with n = 0:
U0+1 =
Next we use n = 1:
U1+1 =
U1+1 =
Note the use of parentheses. It is an extremely good habit to use parentheses when replacing one expression with another one. With the parentheses it is clear that we must use the Distributive Property to multiply correctly:
b. Find a if and a > 0:
Using the equation we got for we can replace with 30:
Now we can solve for a. This is a quadratic equation so we want one side to be zero. Subtracting 30 from each side we get:
Now we either factor or use the Quadratic Formula. This factors easily:
(a+6)(a-3) = 0
From the Zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
a+6 = 0 or a-3 = 0
Solving these we get:
a = -6 or a = 3
Since we were told that a > 0 we must reject the a = -6 solution. So a = 3.
You can put this solution on YOUR website!
a. U_0 = 16 ===> . Then .
b.If , then , and (a+6)(a-3) = 0. Therefore a = 3. (Eliminate -6 because we want a>0).
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