This is out of the OG11 DS#129
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
i didn't understand the explanation in the OG book, someone please elaborate on the answer?? why is A correct?
thanks in advance
arithmetic operations - exponents help!
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The question being asked is - "Is b-a >= 3^n - 2^n.
From stmt 1, it is given a = 2^(n+1) and b = 3^(n+1)
==> b - a will be 3^(n+1) - 2^(n+1)
which can be written as 3.3^n - 2.2^n (x^(n+1) = x*x^n)
This can further be written as (3^n + 2.3^n) - 2.2^n ==> 3^n + 2(3^n - 2^n)
b - 1 = 3^n + 2(3^n - 2^n)
From here is it clear that b-a is greater than (3^n - 2^n). Hence sufficient.
From smtt2, n = 3 but no information is given abt a and b. Hence Insufficient.
IMO A.
From stmt 1, it is given a = 2^(n+1) and b = 3^(n+1)
==> b - a will be 3^(n+1) - 2^(n+1)
which can be written as 3.3^n - 2.2^n (x^(n+1) = x*x^n)
This can further be written as (3^n + 2.3^n) - 2.2^n ==> 3^n + 2(3^n - 2^n)
b - 1 = 3^n + 2(3^n - 2^n)
From here is it clear that b-a is greater than (3^n - 2^n). Hence sufficient.
From smtt2, n = 3 but no information is given abt a and b. Hence Insufficient.
IMO A.