Question

Using binomial theorem, expand \[\{ {(x + y)^5} + {(x - y)^5}\} \] and hence find the value of \[\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} \].

Answer 1

Hint: To solve this question first thing we should know that ${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}} \cdot {y^r}} $ and here $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, ${\left( {x - y} \right)^n} = {\left( { - 1} \right)^r}\sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}} \cdot {y^r}} $ and on further expanding the equation this question can be solved easily.

Complete step by step answer:
Given, the expression is \[\{ {(x + y)^5} + {(x - y)^5}\} \].
Now, substitute 5 for n, in the equation ${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}} \cdot {y^r}} $.
So,
${\left( {x + y} \right)^5} = \sum\limits_{r = 0}^5 {^5{C_r}{x^{5 - r}} \cdot {y^r}} $, on expanding we get.
\[
  {\left( {x + y} \right)^5}{ = ^5}{C_0}{x^5} \cdot {y^0}{ + ^5}{C_1}{x^4} \cdot y{ + ^5}{C_2}{x^3} \cdot {y^2}{ + ^5}{C_3}{x^2} \cdot {y^3}{ + ^5}{C_4}x \cdot {y^4}{ + ^5}{C_5}{x^0} \cdot {y^5} \\
  {\left( {x + y} \right)^5}{ = ^5}{C_0}{x^5}{ + ^5}{C_1}{x^4} \cdot y{ + ^5}{C_2}{x^3} \cdot {y^2}{ + ^5}{C_3}{x^2} \cdot {y^3}{ + ^5}{C_4}x \cdot {y^4}{ + ^5}{C_5}{y^5} \\
 \]………..(i)
Now, consider the expression, ${\left( {x - y} \right)^n} = {\left( { - 1} \right)^r}\sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}} \cdot {y^r}} $, substitute 5 for n, in the equation ${\left( {x - y} \right)^n} = {\left( { - 1} \right)^r}\sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}} \cdot {y^r}} $.
So,
${ \Rightarrow \left( {x - y} \right)^5} = {\left( { - 1} \right)^r}\sum\limits_{r = 0}^5 {^5{C_r}{x^{5 - r}} \cdot {y^r}} $, on expanding we will get the equation as,
\[
  {\Rightarrow \left( {x - y} \right)^5} = {\left( { - 1} \right)^0}{\,^5}{C_0}{x^5} \cdot {y^0} + {\left( { - 1} \right)^1}{\,^5}{C_1}{x^4} \cdot y + {\left( { - 1} \right)^2}{\,^5}{C_2}{x^3} \cdot {y^2} + {\left( { - 1} \right)^3}{\,^5}{C_3}{x^2} \cdot {y^3} + {\left( { - 1} \right)^1}{\,^5}{C_4}x \cdot {y^4} + {\left( { - 1} \right)^1}{\,^5}{C_5}{x^0} \cdot {y^5} \\
  {\Rightarrow \left( {x - y} \right)^5}{ = ^5}{C_0}{x^5} \cdot {y^0}{ - ^5}{C_1}{x^4} \cdot y{ + ^5}{C_2}{x^3} \cdot {y^2}{ - ^5}{C_3}{x^2} \cdot {y^3}{ + ^5}{C_4}x \cdot {y^4}{ - ^5}{C_5}{x^0} \cdot {y^5} \\
 \]\[\Rightarrow {\left( {x - y} \right)^5}{ = ^5}{C_0}{x^5}{ - ^5}{C_1}{x^4} \cdot y{ + ^5}{C_2}{x^3} \cdot {y^2}{ - ^5}{C_3}{x^2} \cdot {y^3}{ + ^5}{C_4}x \cdot {y^4}{ - ^5}{C_5}{y^5}\]……………………….(ii)

Now, add the equation (i) and (ii).
\[
  {\left( {x + y} \right)^5}{ = ^5}{C_0}{x^5}{ + ^5}{C_1}{x^4} \cdot y{ + ^5}{C_2}{x^3} \cdot {y^2}{ + ^5}{C_3}{x^2} \cdot {y^3}{ + ^5}{C_4}x \cdot {y^4}{ + ^5}{C_5}{y^5} \\
  \underline {{{\left( {x - y} \right)}^5}{ = ^5}{C_0}{x^5}{ - ^5}{C_1}{x^4} \cdot y{ + ^5}{C_2}{x^3} \cdot {y^2}{ - ^5}{C_3}{x^2} \cdot {y^3}{ + ^5}{C_4}x \cdot {y^4}{ - ^5}{C_5}{y^5}} \\
  {\left( {x + y} \right)^5} + {\left( {x - y} \right)^5} = 2\left( {^5{C_0}{x^5}{ + ^5}{C_2}{x^3} \cdot {y^2}{ + ^5}{C_4}x \cdot {y^4}} \right) \\
  {\left( {x + y} \right)^5} + {\left( {x - y} \right)^5} = 2x\left( {^5{C_0}{x^4}{ + ^5}{C_2}{x^2} \cdot {y^2}{ + ^5}{C_4}{y^4}} \right) \\
 \]

Therefore, \[\{ {(x + y)^5} + {(x - y)^5}\} \] is equal to \[2x\left( {^5{C_0}{x^4}{ + ^5}{C_2}{x^2} \cdot {y^2}{ + ^5}{C_4}{y^4}} \right)\], which can be further simplified as follows,
\[
\Rightarrow \{ {(x + y)^5} + {(x - y)^5}\} = 2x\left( {^5{C_0}{x^4}{ + ^5}{C_2}{x^2} \cdot {y^2}{ + ^5}{C_4}{y^4}} \right) \\
\Rightarrow \{ {(x + y)^5} + {(x - y)^5}\} = 2x\left( {\dfrac{{5!}}{{0!\left( {5 - 0} \right)!}}{x^4} + \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}{x^2}{y^2} + \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}}{y^4}} \right) \\
\Rightarrow \{ {(x + y)^5} + {(x - y)^5}\} = 2x\left( {\dfrac{{5!}}{{5!}}{x^4} + \dfrac{{5!}}{{2!3!}}{x^2}{y^2} + \dfrac{{5!}}{{4!1!}}{y^4}} \right) \\
\Rightarrow \{ {(x + y)^5} + {(x - y)^5}\} = 2x\left( {{x^4} + \dfrac{{5 \times 4 \times 3!}}{{\left( {2 \times 1} \right)3!}}{x^2}{y^2} + \dfrac{{5 \times 4!}}{{4!1!}}{y^4}} \right) \\
\Rightarrow \{ {(x + y)^5} + {(x - y)^5}\} = 2x\left( {{x^4} + 10{x^2}{y^2} + 5{y^4}} \right) \\
 \]
Therefore, \[\{ {(x + y)^5} + {(x - y)^5}\} \] is equal to \[2x\left( {{x^4} + 10{x^2}{y^2} + 5{y^4}} \right)\].

Now, we have to find the value of \[\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} \].
If we compare, \[\{ {(x + y)^5} + {(x - y)^5}\} \] and \[\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} \], we can easily find that value of \[\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} \].
As, the value of \[\{ {(x + y)^5} + {(x - y)^5}\} \] is \[2x\left( {^5{C_0}{x^4}{ + ^5}{C_2}{x^2} \cdot {y^2}{ + ^5}{C_4}{y^4}} \right)\].
So, \[\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} \] will be \[2\left( {\sqrt 2 } \right)\left( {^5{C_0}{{\sqrt 2 }^4}{ + ^5}{C_2}{{\sqrt 2 }^2} \cdot {1^2}{ + ^5}{C_4}{1^4}} \right)\], which can be further simplified as follows,
\[
\Rightarrow \{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} = 2\left( {\sqrt 2 } \right)\left( {^5{C_0}{{\sqrt 2 }^4}{ + ^5}{C_2}{{\left( {\sqrt 2 } \right)}^2} \cdot {1^2}{ + ^5}{C_4}{1^4}} \right) \\
\Rightarrow \{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} = 2\left( {\sqrt 2 } \right)\left( {\dfrac{{5!}}{{0!\left( {5 - 0} \right)!}}{{\left( {\sqrt 2 } \right)}^4} + \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}{{\left( {\sqrt 2 } \right)}^2} + \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}}} \right) \\
\Rightarrow \{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} = 2\left( {\sqrt 2 } \right)\left( {\dfrac{{5!}}{{5!}}\left( 4 \right) + \dfrac{{5!}}{{2!3!}}\left( 2 \right) + \dfrac{{5!}}{{4!1!}}} \right) \\
\Rightarrow \{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} = 2\left( {\sqrt 2 } \right)\left( {4 + \dfrac{{5 \times 4 \times 3!}}{{\left( {2 \times 1} \right)3!}}\left( 2 \right) + \dfrac{{5 \times 4!}}{{4!1!}}} \right) \\
\Rightarrow \{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} = 2\left( {\sqrt 2 } \right)\left( {4 + 10\left( 2 \right) + 5} \right) \\
\Rightarrow \{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} = 58\sqrt 2 \\
 \]
Therefore, \[\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\} \] is equal to \[58\sqrt 2 \].

Note: We would like to be able to extend it when a binomial expression is elevated to a power 'n'. In doing this, the binomial theorem supports us. It transforms a phrase like that into a sequence.
The theorem that specifies the expansion of any power \[{\left( {a + b} \right)^m}\] of a binomial \[\left({a+b}\right)\] as a certain sum of products ${a_i}{b_j}$, such as \[{\left( {a + b} \right)^2}={a^2}+2ab+{b^2}\].