sequence & mean

a^n +b^n / a^n-1 + b^n-1 = a + b /2

a^n+b^n/a^n / a^-1+b^nb^-1 = a+b/2

cross multiply
2 (a^n+b^n)=(a+b)(a^na^-1+b^nb^-1)
factor out
2 (a^n+b^n)=(a+b)(a+b)(a^n+b^n)
a^n+b^n=a^2+b^2
n=2

could this be?

but the correct answer is 1

yaa
IMO =1

uff took a while to factorize it

last but one step of the factorizing process is

[b*a^(n+1)]+[a*b^(n+1)]=[a^2*b^n]+[b^2*a^n]

so condition is satisfied only when n=1

HTH

vishu

m not able to factorize it correctly still

UFF ! wil try and scan the working layout ... its way too much to type I will get back on this sooon and yes nice question ! it was fun doing it

Vishu

Where is this question from? The factoring is incredibly complicated and the question itself needs more info to be answerable.

Here's the factoring:

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

Which is only true for ALL a and b if n - 1 = 0 or n = 1.

However, the question doesn't mention all a and b or tell us that a doesn't equal b. So, if we chose a=3 and b=3, then the two sides would be equal for all values of n. If we chose a=3 and b=-3, then the two sides would be equal for all odd values of n.

If the question told us that |a| != |b|, then 1 would be the correct answer. In future, please post all the answer choices.

Thanks Stuart and vishu for help.

Stuart i was getting stuck with factorization, thanks again
this question was from a chapter i was doing on sequence and it did not have multiple answer choices.