KINDLY GIVE ANSWER ALONG WITH PROOF
To evaluate the integral: \[\int \prod_{r=0}^{m} \frac{1}{x + r} \, dx\] we can proceed with the following steps: Step 1: Express the Product as a SumThe integrand is a product of terms of the form \(\frac{1}{x + r}\). To simplify the integration, we can use partial fraction decomposition. Assume thRead more
To evaluate the integral:
\[
\int \prod_{r=0}^{m} \frac{1}{x + r} \, dx
\]
we can proceed with the following steps:
Step 1: Express the Product as a Sum
The integrand is a product of terms of the form \(\frac{1}{x + r}\). To simplify the integration, we can use partial fraction decomposition. Assume that:
\[
\prod_{r=0}^{m} \frac{1}{x + r} = \sum_{r=0}^{m} \frac{A_r}{x + r}
\]
where \(A_r\) are constants to be determined.
Step 2: Determine the Constants \(A_r\)
Multiply both sides by \(\prod_{r=0}^{m} (x + r)\):
\[
1 = \sum_{r=0}^{m} A_r \prod_{\substack{k=0 \\ k \neq r}}^{m} (x + k)
\]
To find \(A_r\), set \(x = -r\). This eliminates all terms in the sum except the one corresponding to \(A_r\):
\[
1 = A_r \prod_{\substack{k=0 \\ k \neq r}}^{m} (-r + k)
\]
Simplify the product:
\[
A_r = \frac{1}{\prod_{\substack{k=0 \\ k \neq r}}^{m} (k – r)}
\]
This can be written as:
\[
A_r = \frac{(-1)^r}{r! (m – r)!}
\]
Step 3: Integrate Term by Term
Now, the integral becomes:
\[
\int \sum_{r=0}^{m} \frac{A_r}{x + r} \, dx = \sum_{r=0}^{m} A_r \int \frac{1}{x + r} \, dx
\]
The integral of \(\frac{1}{x + r}\) is \(\ln|x + r|\), so:
\[
\sum_{r=0}^{m} A_r \ln|x + r| + C
\]
Substitute \(A_r\):
\[
\sum_{r=0}^{m} \frac{(-1)^r}{r! (m – r)!} \ln|x + r| + C
\]
Step 4: Simplify the Expression
The sum can be written in terms of binomial coefficients:
\[
\sum_{r=0}^{m} \frac{(-1)^r}{r! (m – r)!} \ln|x + r| = \frac{1}{m!} \sum_{r=0}^{m} (-1)^r \binom{m}{r} \ln|x + r|
\]
Thus, the final result is:
\[
\boxed{\frac{1}{m!} \sum_{r=0}^{m} (-1)^r \binom{m}{r} \ln|x + r| + C}
\]
Problem: There are 1000 doors, all initially closed. 1000 people walk by these doors. The first person opens every door. The second person toggles every second door (i.e., closes doors 2, 4, 6, etc.). The third person toggles every third door (i.e., doors 3, 6, 9, etc.), and so on. The 1000th personRead more
Problem:
There are 1000 doors, all initially closed. 1000 people walk by these doors. The first person opens every door. The second person toggles every second door (i.e., closes doors 2, 4, 6, etc.). The third person toggles every third door (i.e., doors 3, 6, 9, etc.), and so on. The 1000th person only toggles door 1000. After all the people have passed, how many doors remain open?
Step 1: Understand the pattern
Each person toggles the state of doors that are multiples of their own number. For example, person 12 will toggle doors 12, 24, 36, and so on. The state of each door will change every time it is toggled.
Step 2: Think about how many times each door is toggled
Take any door number, say door 6. It gets toggled by person 1 (since 1 divides 6), person 2 (2 divides 6), person 3 (3 divides 6), and person 6 (6 divides itself). In general, a door will be toggled once for each of its positive divisors.
Step 3: Determine when a door ends up open
Since all doors start closed, each toggle changes its state. So, if a door is toggled an even number of times, it will end up closed. If it is toggled an odd number of times, it will end up open.
Step 4: Identify which door numbers are toggled an odd number of times
From number theory, we know that most numbers have an even number of divisors because divisors usually come in pairs (like 2 and 3 for 6, since 2×3 = 6). However, perfect squares have an odd number of divisors. For example, 36 has divisors: 1, 2, 3, 4, 6, 9, 12, 18, and 36. Notice that 6×6 = 36, so the factor 6 appears only once, not in a pair. This gives it an odd number of total divisors.
Therefore, only the doors with perfect square numbers will remain open at the end.
Step 5: Count the perfect squares between 1 and 1000
The perfect squares less than or equal to 1000 are:
1² = 1,
2² = 4,
3² = 9,
…
31² = 961.
So, there are 31 perfect squares between 1 and 1000.
Final Answer:
See less31 doors will remain open.