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## Here's the Solution to this Question

2.

A.

$P=\infin$ , since the number of negative odd integers is infinite.

B.

$P=999$

C.

$P=\infin$ , since there are infinitely many numbers for denominator

3.

$(xlogx)^2-4\le n^2+4n^2=5n^2\le 5n^4$

so, $f(x)=O(x^4)$

$5x^5\ge Cx^4$ , where C is a constant

so, $g(x)\neq O(x^4)$

1.

$a_0-a_1=a_1-a_2=a_2-a_3=...=a_{n-1}-a_n=1$

$a_0-a_1+a_1-a_2+a_2-a_3+...+a_{n-1}-a_n=n$

$a_0-a_n=n\implies 1-a_n=n$

$a_n=1-n$

proof by induction:

for n=1:

$a_1=a_0-1=0$

$a_1=1-1=0$

let for n=k:

$a_k=1-k=a_{k-1}-1\implies a_k-a_{k-1}=-1$

then for n=k+1:

$a_{k+1}=a_k-1$

$a_{k+1}=1-(k+1)=-k=a_{k-1}-2=a_k+1-2=a_k-1$

2.

A.

$P=\infin$ , since the number of negative odd integers is infinite.

B.

$P=999$

C.

$P=\infin$ , since there are infinitely many numbers for denominator

3.

$(xlogx)^2-4\le n^2+4n^2=5n^2\le 5n^4$

so, $f(x)=O(x^4)$

$5x^5\ge Cx^4$ , where C is a constant

so, $g(x)\neq O(x^4)$

1.

$a_0-a_1=a_1-a_2=a_2-a_3=...=a_{n-1}-a_n=1$

$a_0-a_1+a_1-a_2+a_2-a_3+...+a_{n-1}-a_n=n$

$a_0-a_n=n\implies 1-a_n=n$

$a_n=1-n$

proof by induction:

for n=1:

$a_1=a_0-1=0$

$a_1=1-1=0$

let for n=k:

$a_k=1-k=a_{k-1}-1\implies a_k-a_{k-1}=-1$

then for n=k+1:

$a_{k+1}=a_k-1$

$a_{k+1}=1-(k+1)=-k=a_{k-1}-2=a_k+1-2=a_k-1$

2.

A.

$P=\infin$ , since the number of negative odd integers is infinite.

B.

$P=999$

C.

$P=\infin$ , since there are infinitely many numbers for denominator

3.

$(xlogx)^2-4\le n^2+4n^2=5n^2\le 5n^4$

so, $f(x)=O(x^4)$

$5x^5\ge Cx^4$ , where C is a constant

so, $g(x)\neq O(x^4)$

1.

$a_0-a_1=a_1-a_2=a_2-a_3=...=a_{n-1}-a_n=1$

$a_0-a_1+a_1-a_2+a_2-a_3+...+a_{n-1}-a_n=n$

$a_0-a_n=n\implies 1-a_n=n$

$a_n=1-n$

proof by induction:

for n=1:

$a_1=a_0-1=0$

$a_1=1-1=0$

let for n=k:

$a_k=1-k=a_{k-1}-1\implies a_k-a_{k-1}=-1$

then for n=k+1:

$a_{k+1}=a_k-1$

$a_{k+1}=1-(k+1)=-k=a_{k-1}-2=a_k+1-2=a_k-1$

2.

A.

$P=\infin$ , since the number of negative odd integers is infinite.

B.

$P=999$

C.

$P=\infin$ , since there are infinitely many numbers for denominator

3.

$(xlogx)^2-4\le n^2+4n^2=5n^2\le 5n^4$

so, $f(x)=O(x^4)$

$5x^5\ge Cx^4$ , where C is a constant

so, $g(x)\neq O(x^4)$

1.

$a_0-a_1=a_1-a_2=a_2-a_3=...=a_{n-1}-a_n=1$

$a_0-a_1+a_1-a_2+a_2-a_3+...+a_{n-1}-a_n=n$

$a_0-a_n=n\implies 1-a_n=n$

$a_n=1-n$

proof by induction:

for n=1:

$a_1=a_0-1=0$

$a_1=1-1=0$

let for n=k:

$a_k=1-k=a_{k-1}-1\implies a_k-a_{k-1}=-1$

then for n=k+1:

$a_{k+1}=a_k-1$

$a_{k+1}=1-(k+1)=-k=a_{k-1}-2=a_k+1-2=a_k-1$

2.

A.

$P=\infin$ , since the number of negative odd integers is infinite.

B.

$P=999$

C.

$P=\infin$ , since there are infinitely many numbers for denominator

3.

$(xlogx)^2-4\le n^2+4n^2=5n^2\le 5n^4$

so, $f(x)=O(x^4)$

$5x^5\ge Cx^4$ , where C is a constant

so, $g(x)\neq O(x^4)$

1.

$a_0-a_1=a_1-a_2=a_2-a_3=...=a_{n-1}-a_n=1$

$a_0-a_1+a_1-a_2+a_2-a_3+...+a_{n-1}-a_n=n$

$a_0-a_n=n\implies 1-a_n=n$

$a_n=1-n$

proof by induction:

for n=1:

$a_1=a_0-1=0$

$a_1=1-1=0$

let for n=k:

$a_k=1-k=a_{k-1}-1\implies a_k-a_{k-1}=-1$

then for n=k+1:

$a_{k+1}=a_k-1$

$a_{k+1}=1-(k+1)=-k=a_{k-1}-2=a_k+1-2=a_k-1$

v