Question

Weak Induction

Answer 1

Answer:

Following are the answer to this question:

Explanation:

In option 1:

The value of n is= 7, which is (base case)

$\to 3^7$

when n=k for the true condition:

$\to 3^k$

when n=k+1 it tests the value:

$\to 3^{(k+1)}= 3^k,3\\\to < (k!) 3 \ substituting \ equation \\\to$

since k>6  hence the value is KH>3 hence proved.

In option 2:

when:

for n=1:(base case)

$\log(1!)$

0<=0 \\ condition is true

when the above statement holds value n=1

when n=k

$\log(k!)$

when n=k+1

$\log(k+1)!=\log(k!)+\log(k+1)$

             

$\because k \log k$      $[\therefore KH>K \Rightarrow \log(KH>\loK)]$

In option 3:

when n=1:

$A_1 \cup B=A_1 \cup B$

when n=k

$\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a$

hence n=k+1 is true.