Question

Find the value of: ${(2 + \sqrt 5 )^5} + {(2 - \sqrt 5 )^5}$.

Answer 1

Hint: We will use binomial theorem to simplify both the expressions in our question. After that, we will observe some terms are similar and some will cut each other off, due to which calculations become extremely easy.

Complete step-by-step answer:
We know that the binomial theorem states that:
${(a + b)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^{n - 1}}{C_1}{a^n}{b^1} + .......{ + ^n}{C_{n - 1}}{a^1}{b^{n - 1}}{ + ^n}{C_n}{a^0}{b^n}$
Now, we will first just take a = 2, $b = \sqrt 5 $ and n = 5.
Putting these values in formula, we will get:-
${(2 + \sqrt 5 )^5}{ = ^5}{C_0}{\left( 2 \right)^5}{\left( {\sqrt 5 } \right)^0}{ + ^5}{C_1}{\left( 2 \right)^4}{\left( {\sqrt 5 } \right)^1} + .......{ + ^5}{C_4}{\left( 2 \right)^1}{\left( {\sqrt 5 } \right)^4}{ + ^5}{C_5}{\left( 2 \right)^0}{\left( {\sqrt 5 } \right)^5}$ ………..(1)
Now, in the same formula, we will put a = 2, $b = - \sqrt 5 $ and n = 5.
Putting these values, we will then obtain the expression as follows:-
${(2 - \sqrt 5 )^5}{ = ^5}{C_0}{\left( 2 \right)^5}{\left( { - \sqrt 5 } \right)^0}{ + ^5}{C_1}{\left( 2 \right)^4}{\left( { - \sqrt 5 } \right)^1} + .......{ + ^5}{C_4}{\left( 2 \right)^1}{\left( { - \sqrt 5 } \right)^4}{ + ^5}{C_5}{\left( 2 \right)^0}{\left( { - \sqrt 5 } \right)^5}$ ……..(2)
Now, we can clearly observe in (1) and (2), that the terms with even powers of $\sqrt 5 $ or $ - \sqrt 5 $ are replica of each other. Therefore, we can just take twice of either of the term among them while adding both of these equations (1) and (2). The terms with odd powers of $\sqrt 5 $ or $ - \sqrt 5 $ will get cancelled out (one being positive and other being negative) while addition of the equations (1) and (2).
So, addition of the equation (1) and (2) gives us the following expression:-
\[ \Rightarrow {(2 + \sqrt 5 )^5} + {(2 - \sqrt 5 )^5} = 2\left[ {^5{C_0}{{\left( 2 \right)}^5}{{\left( {\sqrt 5 } \right)}^0}{ + ^5}{C_2}{{\left( 2 \right)}^3}{{\left( {\sqrt 5 } \right)}^2}{ + ^5}{C_4}{{\left( 2 \right)}^1}{{\left( {\sqrt 5 } \right)}^4}} \right]\] ……..(3)
Now, we will use the formula: $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
So, we will get:-
$^5{C_0} = \dfrac{{5!}}{{0!(5)!}} = 1{,^5}{C_2} = \dfrac{{5!}}{{2!(3)!}} = 10{,^5}{C_4} = \dfrac{{5!}}{{4!(1)!}} = 5$
Now, putting these in (3), we will get:-
\[ \Rightarrow {(2 + \sqrt 5 )^5} + {(2 - \sqrt 5 )^5} = 2\left[ {{{\left( 2 \right)}^5}{{\left( {\sqrt 5 } \right)}^0} + 10 \times {{\left( 2 \right)}^3}{{\left( {\sqrt 5 } \right)}^2} + 5 \times {{\left( 2 \right)}^1}{{\left( {\sqrt 5 } \right)}^4}} \right]\]
Simplifying the RHS to get as follows:-
\[ \Rightarrow {(2 + \sqrt 5 )^5} + {(2 - \sqrt 5 )^5} = 2\left( {32 + 10 \times 8 \times 5 + 5 \times 2 \times 25} \right)\]
Simplifying the RHS further to get as follows:-
\[ \Rightarrow {(2 + \sqrt 5 )^5} + {(2 - \sqrt 5 )^5} = 2\left( {32 + 400 + 250} \right)\]
Simplifying the values inside the bracket in RHS, we will get:-
\[ \Rightarrow {(2 + \sqrt 5 )^5} + {(2 - \sqrt 5 )^5} = 2 \times 682\]
Simplifying for the last time, we will eventually get:-
\[ \Rightarrow {(2 + \sqrt 5 )^5} + {(2 - \sqrt 5 )^5} = 1364\].

Hence, the required answer is 1364.

Note: The students must note that even if they do not observe the pattern as shown above, they can calculate the values normally as well and will get the same answer only but that will just include too many calculations which may lead to some mistake.
Binomial theorem was one of the ways for selecting k objects from n objects without replacement was of extreme interest for Indian Mathematicians. It was first seen in the fourth century BC when the exponent was taken to be 2.