Problem Solving

( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )


A) a^2^n - b^2^n
B) a ^2n + b^2n
C) (a^n - b^n) ^ 2
D) (a^2 - b^2) ^ (n-1)
E) (a^2 + b^2) ^ (n-1)


OA : A

cramya wrote:( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )


A) a^2^n - b^2^n
B) a ^2n + b^2n
C) (a^n - b^n) ^ 2
D) (a^2 - b^2) ^ (n-1)
E) (a^2 + b^2) ^ (n-1)


OA : A

substitute n=1

( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) ) = (a+b)(a-b) = a^2-b^2



only A satisfies.

cramya wrote:( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )


A) a^2^n - b^2^n
B) a ^2n + b^2n
C) (a^n - b^n) ^ 2
D) (a^2 - b^2) ^ (n-1)
E) (a^2 + b^2) ^ (n-1)


OA : A

Picking numbers is a great way to go. We could also solve this in about 15 seconds by recognizing that we have a difference of squares and peeking at the answer choices.

Any quadratic in the form of:

(x + y) (x - y)

can be rewritten as:

x^2 - y^2

Once we see that the original expression is a difference of squares, we should quickly spot that only (A) follows the form that we want.

Down with algebra! Up with critical thinking!

I have a doubt.
I recognised the pattern but I also tried to quickly solve.

a^2^(n-1) can be written as a^(2*(n-1))

Using this, I arrive at a^(4*(n-1)) - b^(4*(n-1)) and hence (a^4n)/(a^4) - (b^4n)/(b^4)
How do I use this to arrive at the answer format?
I know that this is not the most time-conscious way to solve, but I was just a little curious.

Stuart,
My bad for not including source.Its from one of the quantitative aptitude books.

My approach was this and please correct me if I am mistaken. I felt picking numbers was may be easier like u said than the algebric approach below. Posted it to point out the application of the exponent rules.


( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

Of the form (a+b) (a-b) = a^2 - b^2

a = a^2^(n-1) and b = b^2^(n-1)



( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

= (a^2^(n-1)) ^ 2 - (b^2^(n-1))^2


I then interpreted this (x^y)^z = x ^ (y*z)

where x= a ; y = 2^(n-1) ; z=2 and x=b;y = 2^(n-1) ; z=2

a ^(2^(n-1) * 2) - b ^ (2^(n-1) * 2)

= a ^(2^(n-1) * 2^1) - b ^ (2^(n-1) * 2^1)

{Using x^c * x^d = x^(c+d) where x=2,c=n-1 d=1}


= a ^ (2^n) - b ^ ( 2^n)

= a^2^n - b^2^n



I tried to represent it as unambigous as possible and in that process may have missed what I wanted to achieve.

Also my thought was if its given

as a^b^c then this is not equal to a ^(b*c)

Only (a^b)^c = a ^ (b*c)


There was also a GMAT prep that read (may not be the exact prob but just to illustrate) 2^3^4 where we had to 3^4 first then raise 2 to that power

2^3^4 is not equal to 2 ^(3*4)


Sorry for the long post....


Regards,
Cramya

2(n-1)'s are distracting - the question stem is actually

(a^x+b^X)(a^x-b^x) which is equal to

a^2x-b^2x

I am not mistaken this is equal to: a^(4(n-1))-b^(4(n-1))

cramya wrote:
( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

Of the form (a+b) (a-b) = a^2 - b^2

a = a^2^(n-1) and b = b^2^(n-1)



( a^2^(n-1) + b^2^(n-1) ) ( a^2^(n-1) - b^2^(n-1) )

= (a^2^(n-1)) ^ 2 - (b^2^(n-1))^2


I then interpreted this (x^y)^z = x ^ (y*z)

where x= a ; y = 2^(n-1) ; z=2 and x=b;y = 2^(n-1) ; z=2

a ^(2^(n-1) * 2) - b ^ (2^(n-1) * 2)

= a ^(2^(n-1) * 2^1) - b ^ (2^(n-1) * 2^1)

{Using x^c * x^d = x^(c+d) where x=2,c=n-1 d=1}


= a ^ (2^n) - b ^ ( 2^n)

= a^2^n - b^2^n

I could not help, but acknowledge an excellent way to solve the problem besides choosing the number!

Thanks cramya.