Question

If${{(6\sqrt{6}+14)}^{2n+1}}=P$, prove that the integral part of P is an even integer and PF= ${{20}^{n+1}}$where F is the fractional part of P.

Answer 1

Hint: Let the fractional part of the equation which is F =${{(6\sqrt{6}-14)}^{2n+1}}$.
To prove that integral part of P is an even integer we need to solve P-F.

Formula Used:
${{(a-b)}^{n}}{{(a+b)}^{n}}={{({{a}^{2}}-{{b}^{2}})}^{n}}$,
\[\] $nCr=C(n,r)=n!/\text{(n-r)}!\text{r}!$
And
${{(x+y)}^{n}}=\sum\limits_{k=0}^{n}{(n!/(n-k)}!r!){{x}^{n-k}}{{y}^{k}}$

Complete step-by-step answer:
 it is given that,
${{(6\sqrt{6}+14)}^{2n+1}}=P$
F is the fractional part which is
= ${{(6\sqrt{6}-14)}^{2n+1}}$
Now,
$6\sqrt{6}-14=\{{{(6\sqrt{6})}^{2}}-{{14}^{2}}\}/(6\sqrt{6}+14)$
=$(216-196)/(6\sqrt{6}+14)$
=$20/(6\sqrt{6}+14)$
< 1
Now,
We have already
Let F=${{(6\sqrt{6}-14)}^{2n+1}}$
Now,
\[\begin{align}
  & P-F={{[6\sqrt{6}~+\text{ }14]}^{2n+1}}-{{[\text{ }6\sqrt{6}~-\text{ }14]}^{2n+1}} \\
 & =\text{ }{{(6\sqrt{6})}^{2n+1}}{{+}^{2n+1}}C1{{(6\sqrt{6})}^{2n}}.14\text{ }+\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots .{{+}^{2n+1}}C2n+{{114}^{2n+1}}~-\text{ }\{{{(6\sqrt{6})}^{2n+1}} \\
 & {{-}^{2n+1}}C1{{(6\sqrt{6})}^{2n}}.14\text{ }+\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots .{{-}^{2n+1}}C2n+{{114}^{2n+1}}\} \\
\end{align}\]
Here (6$\sqrt{6}$)\[^{2n+1}\] will cancel out with each other
And we will be left with
\[2{{[}^{2n+1}}C1{{(6\sqrt{6})}^{2n}}.14\text{ }+\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots .{{+}^{2n+1}}C2n+{{114}^{2n+1}}]\]
$\Rightarrow $ P-F is an integer. Since, the common multiple is 2
$\Rightarrow $ it is even
Hence, P-F is an even integer.
Now,
PF=${{(6\sqrt{6}+14)}^{2n+1}}$${{(6\sqrt{6}-14)}^{2n+1}}$
{Using the formula ${{(a-b)}^{n}}{{(a+b)}^{n}}={{({{a}^{2}}-{{b}^{2}})}^{n}}$}
 \[\begin{align}
  & ={{[{{(6\sqrt{6})}^{2}}-\text{ }{{\left( 14 \right)}^{2}}]}^{2n+1}} \\
 & =\text{ }{{\left( 216-196 \right)}^{2n+1}} \\
 & =\text{ }{{20}^{2n+1}} \\
\end{align}\]

Additional Information:
The largest integer that does not exceed x is called the integral part of the number x (designated by [x]). Thus, [5.6] = 5, [-3.2] = 4.
The function [x] is called the largest integer function. The difference {x} = x- [x] is called the fractional part of x (designated by {x}).
The following always holds: 0 ^ {x} < 1. The function {x} is a periodic function with a period one. Closely connected with the fractional part of a number is the concept of the distance to the nearest integer x
[represented by (x)], which is defined as follows:
(x)=min [x—k] k=0, ±1,\[\pm 2,~\ldots \] \[\]

Note: Here we are trying to find the integral part by letting the fractional part of the given value of P and we are using the binomial theorem to solve the given value of P as it is in the form of binomial expression.
Some points to remember:
The total number of terms in the expansion of\[{{\left( x+y \right)}^{n}}\] are (n+1)
The sum of exponents of x and y is always n.
\[n{{C}_{0}},\text{ }n{{C}_{1}},\text{ }n{{C}_{2}},\text{ }\ldots \text{ }..,\text{ }n{{C}_{n}}~\]are called binomial coefficients and also represented by \[{{C}_{0}},\text{ }{{C}_{1}},\text{ }C2,\text{ }\ldots ..,\text{ }{{C}_{n}}\].