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

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

Explanation

Solution

To solve this question first thing we should know that (x+y)n=∑r=0nnCrxn−r⋅yr{\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}} \cdot {y^r}} and here nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, (x−y)n=(−1)r∑r=0nnCrxn−r⋅yr{\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\\{ {(x + y)^5} + {(x - y)^5}\\} .
Now, substitute 5 for n, in the equation (x+y)n=∑r=0nnCrxn−r⋅yr{\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}} \cdot {y^r}} .
So,
(x+y)5=∑r=055Crx5−r⋅yr{\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} \\
$$⇒(x−y)5=5C0x5−5C1x4⋅y+5C2x3⋅y2−5C3x2⋅y3+5C4x⋅y4−5C5y5\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).

(x+y)5=5C0x5+5C1x4⋅y+5C2x3⋅y2+5C3x2⋅y3+5C4x⋅y4+5C5y5 (x−y)5=5C0x5−5C1x4⋅y+5C2x3⋅y2−5C3x2⋅y3+5C4x⋅y4−5C5y5‾ (x+y)5+(x−y)5=2(5C0x5+5C2x3⋅y2+5C4x⋅y4) (x+y)5+(x−y)5=2x(5C0x4+5C2x2⋅y2+5C4y4)  {\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\\{ {(x + y)^5} + {(x - y)^5}\\} is equal to 2x(5C0x4+5C2x2⋅y2+5C4y4)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,

⇒(x+y)5+(x−y)5=2x(5C0x4+5C2x2⋅y2+5C4y4) ⇒(x+y)5+(x−y)5=2x(5!0!(5−0)!x4+5!2!(5−2)!x2y2+5!4!(5−4)!y4) ⇒(x+y)5+(x−y)5=2x(5!5!x4+5!2!3!x2y2+5!4!1!y4) ⇒(x+y)5+(x−y)5=2x(x4+5×4×3!(2×1)3!x2y2+5×4!4!1!y4) ⇒(x+y)5+(x−y)5=2x(x4+10x2y2+5y4) \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\\{ {(x + y)^5} + {(x - y)^5}\\} is equal to 2x(x4+10x2y2+5y4)2x\left( {{x^4} + 10{x^2}{y^2} + 5{y^4}} \right).

Now, we have to find the value of (2+1)5+(2−1)5\\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\\} .
If we compare, (x+y)5+(x−y)5\\{ {(x + y)^5} + {(x - y)^5}\\} and (2+1)5+(2−1)5\\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\\} , we can easily find that value of (2+1)5+(2−1)5\\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\\} .
As, the value of (x+y)5+(x−y)5\\{ {(x + y)^5} + {(x - y)^5}\\} is 2x(5C0x4+5C2x2⋅y2+5C4y4)2x\left( {^5{C_0}{x^4}{ + ^5}{C_2}{x^2} \cdot {y^2}{ + ^5}{C_4}{y^4}} \right).
So, (2+1)5+(2−1)5\\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\\} will be 2(2)(5C024+5C222⋅12+5C414)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,

⇒(2+1)5+(2−1)5=2(2)(5C024+5C2(2)2⋅12+5C414) ⇒(2+1)5+(2−1)5=2(2)(5!0!(5−0)!(2)4+5!2!(5−2)!(2)2+5!4!(5−4)!) ⇒(2+1)5+(2−1)5=2(2)(5!5!(4)+5!2!3!(2)+5!4!1!) ⇒(2+1)5+(2−1)5=2(2)(4+5×4×3!(2×1)3!(2)+5×4!4!1!) ⇒(2+1)5+(2−1)5=2(2)(4+10(2)+5) ⇒(2+1)5+(2−1)5=582 \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, (2+1)5+(2−1)5\\{ {(\sqrt 2 + 1)^5} + {(\sqrt 2 - 1)^5}\\} is equal to 58258\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 (a+b)m{\left( {a + b} \right)^m} of a binomial (a+b)\left({a+b}\right) as a certain sum of products aibj{a_i}{b_j}, such as (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2}={a^2}+2ab+{b^2}.