IIT JEE , AIEEE Forum - Mathematics , Algebra

hpudipeddi

(2003+1/2)^n+(2004+1/2)^n is a positive integer when

a)n is even

b)n is odd

c)n=117 or 119

d)n=1 or n=3

e)none of these

(Please give me a detailed answer)

mukundmadhav

Take 1/2^n common

(1/2)^n[4007^n + 4009^n]

(1/2)^n[(4008-1)^n + (4008+1)^n]

(1/2)^n-1 [4008^n + nC2x4008^n-2.... even terms

Now this holds for n=1, n=3 but not for n=5.. Because there you get a term 4008x5.. Which has only 3 powers of 2 in it while the denominator has 4..

So answer is n=1,3

hpudipeddi

I didn't understand the last 2 lines
Please Explain

mukundmadhav

After you obtain the final expression, you place values of n in it.. 1, 3 satisfy
When you put n=5. The third term in the bracket that you get is 5x4008
4008=8X501
And in the denominator you have 2^4
So you'll get a fraction and not an integer

hpudipeddi

Ok i understood but what about 117 and 119

mukundmadhav

was it a multiple answer question?

hpudipeddi

Yes it is a multiple answer question i will give the options once more:
a)n is even

b)n is odd

c)n=1 and n=3

d)n=117 or n=119

e)none of these

mukundmadhav

Hmm... See for all odd numbers greater than 3, the term nx4008 will always be there where n is the number. And in the denominator 2^n-1 will be there and as we've seen 4008 contains only 3 powers of 2..
So.. No other odd numbers after 1 and 3

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