Answer:
Final answer is
.
Step-by-step explanation:
Given expression is
.
Now we need to find the product of the given expression
.



combine like terms


Hence final answer is
.
Answer:
0B
Step-by-step explanation:
Answer:
Check the explanation
Step-by-step explanation:
(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n
(b) (n!)^2=-1 mod p now by format theorem (n!)^(p-1)= 1 mod p ( as p doesn't divide (n!)^2)
Hence (-1)^(p-1)/2= 1 mod p hence [ as p-1/2 is an integer] and hence( p-1)/2 is even number hence p is of the form 4k+1
(C) now let p be the largest prime of the form 4k+1 consider x= (p!)^2+1 . Let q be the smallest prime dividing x . By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite
9514 1404 393
Answer:
M₅ = 1.6570
Step-by-step explanation:
The basic idea is to find the function values at x ∈ {1.1, 1.3, 1.5, 1.7, 1.9}, add those values, and multiply the result by 1/5. Those x-values are the midpoints of intervals 0.2 units wide, between x = 1 and x = 2. The width 0.2 is the total interval width (2-1=1) divided by n = 5.
The calculation and results are shown in the attachment. Rounded to 4 decimal places, the integral is ...
M₅ = 1.6570
__
The actual integral is about 1.6574, so this is pretty close.