Neetesh Kumar | December 31, 2024
Calculus Homework Help
This is the solution to Math 1c
Assignment: 11.1 Question Number 37
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Step-by-step solution:
Part (a): Determine monotonicity:
Step 1: Compute an+1 and compare it to an:
The sequence is an=4n+33n−4.
To determine if it is increasing or decreasing, we analyze the difference an+1−an.
Step 2: Write an+1:
Substitute n+1 into the formula for an:
an+1=4(n+1)+33(n+1)−4=4n+4+33n+3−4=4n+73n−1
Step 3: Analyze an+1−an:
Compute the difference:
an+1−an=4n+73n−1−4n+33n−4
Combine under a common denominator:
an+1−an=(4n+7)(4n+3)(3n−1)(4n+3)−(3n−4)(4n+7)
Expand the numerator:
(3n−1)(4n+3)=12n2+9n−4n−3=12n2+5n−3
(3n−4)(4n+7)=12n2+21n−16n−28=12n2+5n−28
Subtract the two terms:
an+1−an=(4n+7)(4n+3)(12n2+5n−3)−(12n2+5n−28)
Simplify the numerator:
12n2+5n−3−12n2−5n+28=25
Thus:
an+1−an=(4n+7)(4n+3)25
Step 4: Sign of an+1−an:
The denominator (4n+7)(4n+3)>0 for all n≥1, and the numerator 25>0.
Therefore:
an+1−an>0
This shows that the sequence is increasing.
Part (b): Determine boundedness:
Step 1: Check for upper and lower bounds:
The sequence is:
an=4n+33n−4
For large n, divide numerator and denominator by n:
an=4+n33−n4→43as n→∞
For n=1, a1=4(1)+33(1)−4=7−1=−71. Therefore, the sequence is bounded below by −71 and above by 43.
Step 2: Conclusion:
The sequence is bounded.
Final Answers:
(a) The sequence is increasing
(b) The sequence is bounded
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