Respuesta :
Answer:
Following are the answer to this question:
Explanation:
In option 1:
The value of n is= 7, which is (base case)
[tex]\to 3^7<7!\\\to 2187<5050\\[/tex]
when n=k for the true condition:
[tex]\to 3^k<k!......(i)\\\\[/tex]
when n=k+1 it tests the value:
[tex]\to 3^{(k+1)}= 3^k,3\\\to < (k!) 3 \ substituting \ equation \\\to <k! \cdot (k+1)\\[/tex]
since k>6 hence the value is KH>3 hence proved.
In option 2:
when:
for n=1:(base case)
[tex]\log(1!)<=1\log(1)[/tex]
0<=0 \\ condition is true
when the above statement holds value n=1
when n=k
[tex]\log(k!) <=k\log(k)....(1)[/tex]
when n=k+1
[tex]\log(k+1)!=\log(k!)+\log(k+1)\\[/tex]
[tex]<= k \log(k)+\log(kH)\\<= kH\log(kH)\\[/tex]
[tex]\because k \log k<=k log(KH)\\\\[/tex] [tex][\therefore KH>K \Rightarrow \log(KH>\loK)][/tex]
In option 3:
when n=1:
[tex]A_1 \cup B=A_1 \cup B\\\\[/tex]
when n=k
[tex]\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a \\\\[/tex]
hence n=k+1 is true.
