Is the question correct?

I think its b-a

This question is from the OG, and I believe it should be b-a and 2^n in the prompt, not b-1 and 2^2. The question should read:

if n is a positive integer, is the value of b-a at least twice the value of 3^n-2^n?

1.a=2^n+1 and b=3^n+1
2.n=3

The prompt can be re-written as:

b-a >= 2*(3^n - 2^n) ?

b-a >= 2*(3^n) - 2^(n+1) ?

Statement 1:

a=2^n+1 and b=3^n+1

Therefore, using the inequality from the prompt, we want to know if it's true that:

3^(n+1) - 2^(n+1) >= 2*(3^n) - 2^(n+1)

Add 2^(n+1) to both sides:

3^(n+1) >= 2*(3^n)

3*(3^n) >= 2*(3^n)

Since we know 3^n is positive, we know we can divide both sides of the inequality by 3^n without the sign changing.

3 >= 2.

THat is always true, and therefore, St (1) is sufficient.

Statement 2:

Nothing about a or b ... insufficient

thanks raz i was stucked with 2.n=3 which has no meaning

b/a = 3^n+1/2^n+1
= (3.3^n)/(2.2^n)=(3/2)(3^n/2^n)=1.5(3^n/2^n)

now, since n is positive integer n=1,2,3....
for n= 1 the value stands for 1.5*1.5=2.25 which will go up if we set n=2,3,4,...

so answer is A.

Is my method ok?

I got the answer A.... which is correct. But my method was plugging in a value for n. To make things simple, I made n=1

Would my method be correct?


Thanks

its notb/a its b-a