I'm partial to solving with generating functions. Let
Multiply both sides of the recurrence by 
and sum over all 
.
Shift the indices and factor out powers of 
as needed so that each series starts at the same index and power of .
Now we can write each series in terms of the generating function 
. Pull out the first few terms so that each series starts at the same index 
.
Solve for :
Splitting into partial fractions gives
which we can write as geometric series,
which tells us
# # #
Just to illustrate another method you could consider, you can write the second recurrence in matrix form as
By substitution, you can show that
or
Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of 
, then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.
Answer:
c. <N
Step-by-step explanation:
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20(15x − 34)+9x
Multiply the bracket by 20
(20)(15x)=300x
(20)(-34)=-680
300x-680+9x
300x+9x-680 ( combine like terms)
Answer: 309x-680
Answer:
645;lllll
Step-by-step explanation:
Answer:
See below
Step-by-step explanation:
Quadratic formula :
x = [-b ± sqrt ( b^2 - 4ac) ] / 2a
for this problem a = 3 b= - 5 c = - 4
substitute in the values and compute:
<u> x= 5/6 ± sqrt (73)/6 </u>
