Sequence And Series

We are asked to find the value of the sum:

First, factor the denominator:

So, the general term in the sum is:

Now, apply partial fraction decomposition to :

Multiplying both sides by gives:

Expanding and solving for and :

Equating coefficients, we get:

and

and

Thus, we have:

Now, we can rewrite the sum as a telescoping series:

Writing out the terms, we see most of the terms cancel, leaving us with:

Now, subtract:

The answer is .

The -th term of the series is given by .
We can split the series into two parts:
When is odd, , so .
When is even, , so .
Thus, the series alternates between terms of the form for odd and for even .
We now need to find the sum of the first 40 terms. The sum of the terms for odd is the sum of , and for even , it is the sum of .
For odd , the terms are: , up to the 40th term. This gives us a sum of 820.

The correct option is (B) :

We are given that: This can be written as: Using the formula for the sum of squares: Substitute this into the expression for : Now, we need to calculate . For :

The correct option is (B) :

The general form of the \textsuperscript{th} term in a geometric series is given by: where is the first term and is the common ratio. Here, the series is given by: Thus, the first term , and the common ratio . We are asked to find the 11\textsuperscript{th} term, so we apply the formula for the \textsuperscript{th} term: Now, calculate : Thus, the 11\textsuperscript{th} term is:

The correct option is (C) :

We start with the given expression: This can be transformed using the identity to relate and .
Recognizing that and are reciprocal squares, we can apply the identity: Using the pythagorean identities and , we realize: Thus, we see: The values and are always greater than or equal to 1 in the interval , so must be greater than or equal to 2: Given that can grow infinitely as approaches 0 or (where or blow up), indeed lies in the interval .

The period of is given by .
Here, , so the period of is: Thus, the correct answer is .

To find the domain of , we need to ensure that the argument of the inverse sine function, , lies within the interval and that the denominator, , remains non-zero and real.
1. Argument of within : - The condition simplifies to:
2. Denominator non-zero and real: - The square root is defined and non-zero when:
- This translates to:
3. Intersection of Conditions: - The intersection of and is:
4. Conclusion: - Therefore, the domain of is , where starts at 2 and approaches but does not include 3.

Since the points are collinear, their slopes must be equal: Solving for :

In the expansion of , the general term is: For , we have: Setting exponent of to 3: Substituting : Thus, the coefficient of is 15.

Using the binomial theorem, the sum of binomial coefficients from to is:

The given equation is: To solve this, we recognize that this is a standard integral of the form: For our case, and is replaced by , so we use substitution.
After solving, we find that and .
Thus, . Therefore, the correct answer is .

The determinant of the matrix , denoted , can be calculated using the cofactor expansion.
The formula is: Substituting the given values:

Given that , we first recognize that .
Now consider the expression: which can be simplified by letting and , thus: Using the identity , and substituting back: From the earlier substitution for , : Thus, the expression simplifies to .

We need to find the value of within the specified ranges where .
- For , : Setting gives . However, only negative values are applicable in this range, which does not contribute to the positive solution.
- For , : Solving yields: is within the interval and satisfies .
- For , : Solving results in: Only fits this equation, but must be greater than 2 in this range, so is not valid here.
Thus, the correct and only positive value for which under the constraints provided by the piecewise function is .
Therefore, the correct answer is Option D.

We are asked to find , the conditional probability of given . The formula for conditional probability is given by: From the given information: -
-
-
Substitute the given values into the formula for conditional probability: Thus, the correct answer is .
Thus, the correct answer is option (D), .