OG10 DS 218 at least twice the value

[mitzi]

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

1) a=2ⁿ⁺¹ and b = ⁿ⁺¹
2) n=3

I think that 2) is insufficient(I can get it)
With respect to 1)

first I substituted a and be into the given equation.
3^(n+1) - 2^(n+1)
= 3(3^n) - 2(2^n)

second
3(3^n) - 2(2^n) > 2(3^n) - 2(2^n)

I eliminate -2(2^n) from both sides.
3(3^n) > 2(2^n) is left

in OG's explanation, '2(3^n - 2^n) is left. I don't understand how it is left.
Do I have any calculation mistakes? Will you please help me? Thank you.

[nverma]

Originally Posted by mitzi

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

1) a=2ⁿ⁺¹ and b = ⁿ⁺¹
2) n=3

I think that 2) is insufficient(I can get it)
With respect to 1)

first I substituted a and be into the given equation.
3^(n+1) - 2^(n+1)
= 3(3^n) - 2(2^n)

second
3(3^n) - 2(2^n) > 2(3^n) - 2(2^n)

I eliminate -2(2^n) from both sides.
3(3^n) > 2(2^n) is left

in OG's explanation, '2(3^n - 2^n) is left. I don't understand how it is left.
Do I have any calculation mistakes? Will you please help me? Thank you.

Hi Mitzi

IMO A is the answer..

i was not able to understand thy method..but i have proceeded in the following manner..

substituting the values of b and from stat(1)
3^(n+1) - 2^(n+1) > = 2(3^n) - 2(2^n)
3^n. 3 - 2^n .2 >= 2.3^n - 2.2^n

cancel 2^n .2 from both sides..

3^n. 3 - 2.3^n >=0

3^n >=0

for any real value of n 3^n cannot be 0...so definitely as per the question

b-a is more twice the value of 3^n-2^n..

stat 1 is enough..

[mitzi]

Namaste guru nverma,

Thank you sooooooo much! You see that guru can't even understand what ETS says. No wonder I couldn't even understand it.

Your way is much more concise and intuitive.
I got all the way until this point, but I don't know how you got this

3^n. 3 - 2.3^n >=0 (from here)

3^n >=0 (to here)

Can you help me out how you got this?


for any real value of n 3^n cannot be 0...so definitely as per the question

b-a is more twice the value of 3^n-2^n..

[adi_800]

Originally Posted by mitzi

Namaste guru nverma,

That's some change!!

[nverma]

Originally Posted by mitzi

Namaste guru nverma,

Thank you sooooooo much! You see that guru can't even understand what ETS says. No wonder I couldn't even understand it.

Your way is much more concise and intuitive.
I got all the way until this point, but I don't know how you got this

3^n. 3 - 2.3^n >=0 (from here)

3^n >=0 (to here)

Can you help me out how you got this?


for any real value of n 3^n cannot be 0...so definitely as per the question

b-a is more twice the value of 3^n-2^n..

Hi Mitzi..

I'm just a newbee..NO GURU.. :-)
anyways..

3 (3^n) - 2 (3^n) >= 0 ((3m -2m = m, here m is 3^n))
3^n >=0

I hope its clear now..

[atishree]

Originally Posted by mitzi

Namaste guru nverma,

for any real value of n 3^n cannot be 0...so definitely as per the question

b-a is more twice the value of 3^n-2^n..

PLease somebody explain me this point

[STFUniversity]

Statement (1):
just substitute
b-a = (n+1)-(2n+1)
b-a = -n
we know n is positive, so b-a is negative
3^n - 2^n is obviously positive since we know n is positive (3 - 2 = 1, 3*3 - 2*2 = 5, 3*3*3 - 2*2*2 = 19, etc...)
since b-a is negative and 3^n - 2^n is positive, it's clear that b-a isn't twice the value of 3^n - 2^n
sufficent.

statement (2) tells us nothing about b-a
insufficient.

Answer: A


edit... damnit I missed the exponents in the question. I have got to stop copying and pasting these problems in text format.

[atishree]

Originally Posted by mitzi

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

1) a=2ⁿ⁺¹ and b = ⁿ⁺¹
2) n=3

I think that 2) is insufficient(I can get it)
With respect to 1)

first I substituted a and be into the given equation.
3^(n+1) - 2^(n+1)
= 3(3^n) - 2(2^n)

second
3(3^n) - 2(2^n) > 2(3^n) - 2(2^n)

I eliminate -2(2^n) from both sides.
3(3^n) > 2(2^n) is left

in OG's explanation, '2(3^n - 2^n) is left. I don't understand how it is left.
Do I have any calculation mistakes? Will you please help me? Thank you.

ok lets try to solve this

3(3^n) - 2(2^n)
(3^n) + 2(3^n) - 2(2^n) (as x + 2x =3x where x= (3^n))
(3^n) + 2{(3^n) - (2^n)}
as n is positive this quantity above is at least greater than twice (3^n) - (2^n)

was this what the question was asking?