The series as represented in the task content is divergent and hence; diverges.
<h3>What convergence and divergence in series?</h3>
It follows from the task content that the series defined by the summation expression as represented above is to be identified as divergent or convergent.
It is noteworthy to know that;
If a sequence converges, it means that the limit of the sequence exists as n → ∞. Also, if the limit of the sequence as n → ∞ does not exist, Then, such series is said to diverges.
Therefore, by evaluation as the limit of the expression k sin²k / 1 + k³ as k → ∞ is; Undefined and hence does not exist.
Read more on convergence and divergence of series;
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Answer:
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Step-by-step explanation:
Given that M is a polynomial of degree 3.
So, it has three zeros.
Let the polynomial be
M(x) =a(x-p)(x-q)(x-r)
The two zeros of the polynomial are -4 and 4i.
Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.
Then,
M(x)= a{x-(-4)}(x-4i){x-(-4i)}
=a(x+4)(x-4i)(x+4i)
=a(x+4){x²-(4i)²} [ applying the formula (a+b)(a-b)=a²-b²]
=a(x+4)(x²-16i²)
=a(x+4)(x²+16) [∵i² = -1]
=a(x³+4x²+16x+64)
Again given that M(0)= 53.12 . Putting x=0 in the polynomial
53.12 =a(0+4.0+16.0+64)

=0.83
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Answer:
5/12
Step-by-step explanation:
You have to make it a common denomator and then subract until equeal.