Question

Let $R={{\left( 8+3\sqrt{7} \right)}^{20}}$ and [R] = the greatest integer less than or equal to R. Then which of the following is / are true.
(a) [R] is even.
(b) [R] is odd.
(c) $R-[R]=1-\dfrac{1}{{{\left( 8+3\sqrt{7} \right)}^{20}}}$
(d) None of these.

Answer 1

Hint: Express R in terms of an integer and a fraction. Also, express ${{\left( 8-3\sqrt{7} \right)}^{20}}$ in terms of an integer and a fraction. Now use different operations with the equations you get to verify each of the options separately.

Complete step-by-step answer:
It is given that $R={{\left( 8+3\sqrt{7} \right)}^{20}}$ . We can express R as the sum of [R] and a fraction. Let the fractional part of R be g.
$\therefore R=[R]+g={{\left( 8+3\sqrt{7} \right)}^{20}}$
Also, by using the binomial expansion ${{\left( x+y \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}{{.............}^{n}}{{C}_{n}}{{y}^{n}}$ , we get
  $[R]+g={{\left( 8+3\sqrt{7} \right)}^{20}}{{=}^{20}}{{C}_{0}}{{8}^{20}}{{+}^{20}}{{C}_{1}}{{8}^{19}}\times 3\sqrt{7}{{.............}^{20}}{{C}_{20}}{{\left( 3\sqrt{7} \right)}^{20}}..............(i)$
We can also say that ${{\left( 8-3\sqrt{7} \right)}^{20}}$ is a pure fraction by putting the approximate value of root 7. Let its value be f.
$\therefore f={{\left( 8-3\sqrt{7} \right)}^{20}}{{=}^{20}}{{C}_{0}}{{8}^{20}}{{-}^{20}}{{C}_{1}}{{8}^{19}}\times 3\sqrt{7}{{.............}^{20}}{{C}_{20}}{{\left( 3\sqrt{7} \right)}^{20}}..............(ii)$
Now, we can see that all the terms with odd powers of $3\sqrt{7}$ , i.e., the irrational terms of the expansion of ${{\left( 8-3\sqrt{7} \right)}^{20}}$ are negative and magnitude of this irrational terms is equal to the irrational terms present in the expansion of $R={{\left( 8+3\sqrt{7} \right)}^{20}}$ .
So, if we add equations (i) and (ii), all the irrational terms gets cancelled and we get
$[R]+g+f=2\left( ^{20}{{C}_{0}}{{8}^{20}}{{+}^{20}}{{C}_{2}}{{8}^{18}}\times {{\left( 3\sqrt{7} \right)}^{2}}{{.............}^{20}}{{C}_{20}}{{\left( 3\sqrt{7} \right)}^{20}} \right)............(iii)$
Now, as all the terms in the above equation will have the power of $3\sqrt{7}$ to be even, so the right-hand side of the equation is an integer.
As g and f both are fractions, so their sum must be 1 to give an integer, which is necessary for our equation to be true.
$\begin{align}
  & \therefore f+g=1 \\
 & \Rightarrow g=1-f \\
\end{align}$
$\Rightarrow R-[R]=1-{{\left( 8-3\sqrt{7} \right)}^{20}}$
$\Rightarrow R-[R]=1-\dfrac{{{\left( 8-3\sqrt{7} \right)}^{20}}{{\left( 8+3\sqrt{7} \right)}^{20}}}{{{\left( 8+3\sqrt{7} \right)}^{20}}}$
Now using the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ , we get
$R-[R]=1-\dfrac{{{\left( {{8}^{2}}-{{\left( 3\sqrt{7} \right)}^{2}} \right)}^{20}}}{{{\left( 8+3\sqrt{7} \right)}^{20}}}$
$\Rightarrow R-[R]=1-\dfrac{1}{{{\left( 8+3\sqrt{7} \right)}^{20}}}$
Therefore, option (c) is correct.
Now putting g+f=1 in equation (iii), we get
$[R]+1=2\left( ^{20}{{C}_{0}}{{8}^{20}}{{+}^{20}}{{C}_{2}}{{8}^{18}}\times {{\left( 3\sqrt{7} \right)}^{2}}{{.............}^{20}}{{C}_{20}}{{\left( 3\sqrt{7} \right)}^{20}} \right)$
$\Rightarrow [R]=2\left( ^{20}{{C}_{0}}{{8}^{20}}{{+}^{20}}{{C}_{2}}{{8}^{18}}\times {{\left( 3\sqrt{7} \right)}^{2}}{{.............}^{20}}{{C}_{20}}{{\left( 3\sqrt{7} \right)}^{20}} \right)-1$
Letting $^{20}{{C}_{0}}{{8}^{20}}{{+}^{20}}{{C}_{2}}{{8}^{18}}\times {{\left( 3\sqrt{7} \right)}^{2}}{{.............}^{20}}{{C}_{20}}{{\left( 3\sqrt{7} \right)}^{20}}$ to be constant k.
\[\therefore [R]=2k-1\]
As [R] is of 2k-1 form, it is an odd number. Hence, option (b) is also correct.
Therefore, the correct options are (b) and (c).

Note: Be careful about the signs you are using in the binomial expansions and the operation that you are performing between the first two equations that you get. Also, learn the properties related to the greatest integer function and the fractional part function, as they might be useful.