Application of fractional part in expansion

Algebra Level 5

( a + b c ) n = I + f \displaystyle (a+b\sqrt c)^{n}= I+f where I \displaystyle I is integer part and f \displaystyle f is fractional part..

Which of the following are true if a 2 − 1 = b 2 c \displaystyle a^{2}-1=b^{2}c

  1. Value of 1 − f \displaystyle 1-f is I \displaystyle I

  2. Value of 1 − f \displaystyle 1-f is 1 I \displaystyle \frac { 1 }{ I }

  3. Expansion of ( a − b c ) n \displaystyle (a-b\sqrt c)^{n} is 1 − f \displaystyle 1-f .

  4. Value of 1 + f \displaystyle 1+f is I \displaystyle I

  5. Value of 1 + f \displaystyle 1+f is 1 I \displaystyle \frac { 1 }{ I }

  6. Expansion of ( a − b c ) n \displaystyle (a-b\sqrt c)^{n} is 1 + f \displaystyle 1+f .

  7. 1 1 − f − f = I \displaystyle \frac{1}{1-f}-f=I

  8. 1 1 + f − f = I \displaystyle \frac{1}{1+f}-f=I

Submit the answer as the product of the true statements.

For example if 2,3,5 are true submit the answer as 30

0 < a − b c < 1 \displaystyle 0 <a-b\sqrt c<1 , n is a positive integer. a>1, c>0


The answer is 21.

Prince Loomba by Prince Loomba , 21, India

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1 solution

Prince Loomba
Apr 19, 2016

( a + b c ) n = (a+b\sqrt{c})^{n}= ( n 0 ) n \choose 0 a n a^{n} + ( n 1 ) n \choose 1 a n − 1 × b c 1 a^{n-1} \times b\sqrt c^{1} +......... =I+f (fractional part)
( a − b c ) n = (a-b\sqrt{c})^{n}= ( n 0 ) n \choose 0 a n a^{n} - ( n 1 ) n \choose 1 a n − 1 × b c 1 a^{n-1} \times b\sqrt c^{1} +......... =p+g (fractional part)
Adding, ( a + b c ) n (a+b\sqrt{c})^{n} + ( a − b c ) n (a-b\sqrt{c})^{n} = all integer terms (irrational terms cancelled). Note that ( a − b c ) < 1 (a-b\sqrt{c})<1 . Thus all powers are less than 1 or p = 0 p=0 ................... ............................................. I+f+g=integer, I is integer. Thus f+g is integer. 0 < f + g < 2 0<f+g<2 implies that f+g =1 or g = 1 − f g=1-f .............................................. Since a 2 − b 2 c = 1 a^{2}-b^{2}c=1 it is clear that ( a + b c ) n (a+b\sqrt{c})^{n} X ( a − b c ) n (a-b\sqrt{c})^{n} =1............................................................ ( I + f ) × g = 1 (I+f) \times g=1 or ( I + f ) × ( 1 − f ) = 1 (I+f) \times (1-f)=1 or 1 1 − f − f = I \frac{1}{1-f}-f=I ................................................................................................ Thus 3 and 7 are true. Answer =21

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Following corrections must be made to the problem to make it clear and appropriate :-

● First of all, specify that 0 < a − b c < 1 0\,<\,a- b\sqrt c \,<\,1 .

● Note that, you've mentioned a − b c < 1 a-b\sqrt{c}\,<\,1 which more or less indicates that n n is a positive number, but, you need to specify n ϵ Z + n \,\epsilon\ \mathbb{Z^{+}} so that the possibility of n n being a non - intgeral positive real number gets ruled out.

Notice that, we can't say anything about a + b c a+b\sqrt c , in case n n is a non - integral positive real number.

● c > 0 c\,>0 and a > 1 a>1 .

Notice that, we've special cases where your reasoning works, they're:-

c = 0 , a ≠ 1 c=0,a≠1 and b = c = 0 , a ≠ 1 b=c=0,a≠1 .

Aditya Sky - 5 years ago

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Any more corrections?

Prince Loomba - 5 years ago

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Naa. I don't think so.

Aditya Sky - 5 years ago

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