Question

E If ${(4 + \sqrt {15} )^n} = I + f$, where n is an odd natural number, $I$ is an integer and $0 < f < 1$, then
A) I is an odd integer
B) I is an even integer
C) $(I + f)(1 - f) = 1 $
D) $1 - f = {(4 - \sqrt {15} )^n} $

Answer 1

Hint: Expand the RHS using binomial theorem and then find what is the value of I
${f^{'}} = {(4 - \sqrt {15} )^n}$ First, we consider and then we are going to use binomial identities and expand the until n term, after expanding it to n terms, in the binomial form it gets multiplied by 2. So, it is going to be even as the product of 2 is always even. So, the even integer is equal to $I + f + {f^{'}}$ and when we subtract 1, we will get $f + {f^{'}}$. When So, if we subtract integer from integer, we will get integer itself. So, from that we get that $f + {f^{'}}$ is also integer. So, from that we will get that I is an odd integer.

Complete step by step solution:
So, first we expand them using binomial identities.
${\left( {4 + \sqrt {15} } \right)^n} + {(4 - \sqrt {15} )^n} = 2({4^n} + {}^nC{}_2{4^{n - 2}}{(\sqrt {15} )^2} + {}^n{C_4}{4^{n - 4}}{\left( {\sqrt {15} } \right)^4} + ..{}^n{C_n}{4^1}{(\sqrt {15} )^{n - 1}})$
Since the above expression is an integer with no fraction nor irrational numbers involved and also
Since $0 < f < 1$ also does $0 < {f^{'}} < 1$
Which means that $f + {f^{'}} = 1$
From all the statements, we can come to the conclusion that
The expression above is equal to $I + 1$, which refers
${\left( {4 + \sqrt {15} } \right)^n} + {(4 - \sqrt {15} )^n} = I + 1$
From this equation and from the given equation in the question, we get
\[
  {(4 - \sqrt {15} )^n} = 1 - f \\
  (1 - f)(I + f) = {(16 - 15)^n} \\
 \]
Which implies that it’s
$ = 1$
So, finally
$I + 1 = 2({4^n} + {}^nC{}_2{4^{n - 2}}{(\sqrt {15} )^2} + {}^n{C_4}{4^{n - 4}}{\left( {\sqrt {15} } \right)^4} + ..{}^n{C_n}{4^1}{(\sqrt {15} )^{n - 1}})$
It implies that $I + 1$ is an even integer and $I$ is an odd integer.
Hence, we have got our required solution which is that I is an odd integer.

So, the correct answer is Option A,C,D.

Note: We have to always remember the basic number system identities of odd and even numbers, which makes logics easy and also the fractional part of the problem doesn’t exist after expansion. Since their periods lie between zero and one, they can be equalled to one to which they lead to proving the required problem.