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Question: Use the binomial theorem to expand \[{\left( {X + \dfrac{1}{X}} \right)^6}\] ?...

Use the binomial theorem to expand (X+1X)6{\left( {X + \dfrac{1}{X}} \right)^6} ?

Explanation

Solution

Use the concept of binomial expansion and expand the bracket and solve the terms to simplest forms. Use the concept of exponents that when the base is the same powers can be combined to solve the expansion.

  • A binomial expansion helps us to expand expressions of the form (a+b)n{(a + b)^n} through the formula (a+b)n=∑r=0nnCr(a)n−r(b)r{(a + b)^n} = \sum\limits_{r = 0}^n {^n{C_r}{{(a)}^{n - r}}{{(b)}^r}}
  • Formula of combination is given by nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} , where factorial is expanded by the formula n!=n×(n−1)!=n×(n−1)×(n−2)!....=n×(n−1)×(n−2)....3×2×1n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1
  • For any variable or number, when the base is the same and the terms are in multiplication, we can combine the powers of the terms having the same base. ap.aq=ap+q{a^p}.{a^q} = {a^{p + q}}

Complete step by step solution:
We are given the term (X+1X)6{\left( {X + \dfrac{1}{X}} \right)^6} … (1)
Compare the given expansion with general binomial expansion i.e. (a+b)n{(a + b)^n}
Here n=6;a=X;b=1Xn = 6;a = X;b = \dfrac{1}{X}
We use binomial expansion to expand the given term
⇒(X+1X)6=∑r=066Cr(X)6−r(1X)r\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = \sum\limits_{r = 0}^6 {^6{C_r}{{\left( X \right)}^{6 - r}}{{\left( {\dfrac{1}{X}} \right)}^r}}
⇒(X+1X)6=6C0(X)6−0(1X)0+6C1(X)6−1(1X)1+6C2(X)6−2(1X)2+6C3(X)6−3(1X)3\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6}{ = ^6}{C_0}{\left( X \right)^{6 - 0}}{\left( {\dfrac{1}{X}} \right)^0}{ + ^6}{C_1}{\left( X \right)^{6 - 1}}{\left( {\dfrac{1}{X}} \right)^1}{ + ^6}{C_2}{\left( X \right)^{6 - 2}}{\left( {\dfrac{1}{X}} \right)^2}{ + ^6}{C_3}{\left( X \right)^{6 - 3}}{\left( {\dfrac{1}{X}} \right)^3}
+6C4(X)6−4(1X)4+6C5(X)6−5(1X)5+6C6(X)6−6(1X)6{ + ^6}{C_4}{\left( X \right)^{6 - 4}}{\left( {\dfrac{1}{X}} \right)^4}{ + ^6}{C_5}{\left( X \right)^{6 - 5}}{\left( {\dfrac{1}{X}} \right)^5}{ + ^6}{C_6}{\left( X \right)^{6 - 6}}{\left( {\dfrac{1}{X}} \right)^6}
Solve the values in powers
⇒(X+1X)6=6C0(X)6(1X)0+6C1(X)5(1X)1+6C2(X)4(1X)2+6C3(X)3(1X)3\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6}{ = ^6}{C_0}{\left( X \right)^6}{\left( {\dfrac{1}{X}} \right)^0}{ + ^6}{C_1}{\left( X \right)^5}{\left( {\dfrac{1}{X}} \right)^1}{ + ^6}{C_2}{\left( X \right)^4}{\left( {\dfrac{1}{X}} \right)^2}{ + ^6}{C_3}{\left( X \right)^3}{\left( {\dfrac{1}{X}} \right)^3}
+6C4(X)2(1X)4+6C5(X)1(1X)5+6C6(X)0(1X)6{ + ^6}{C_4}{\left( X \right)^2}{\left( {\dfrac{1}{X}} \right)^4}{ + ^6}{C_5}{\left( X \right)^1}{\left( {\dfrac{1}{X}} \right)^5}{ + ^6}{C_6}{\left( X \right)^0}{\left( {\dfrac{1}{X}} \right)^6}
We know that any value with power 0 is equal to 1.
⇒(X+1X)6=6C0(X)6+6C1(X)5(1X)1+6C2(X)4(1X)2+6C3(X)3(1X)3\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6}{ = ^6}{C_0}{\left( X \right)^6}{ + ^6}{C_1}{\left( X \right)^5}{\left( {\dfrac{1}{X}} \right)^1}{ + ^6}{C_2}{\left( X \right)^4}{\left( {\dfrac{1}{X}} \right)^2}{ + ^6}{C_3}{\left( X \right)^3}{\left( {\dfrac{1}{X}} \right)^3}
+6C4(X)2(1X)4+6C5(X)1(1X)5+6C6(1X)6{ + ^6}{C_4}{\left( X \right)^2}{\left( {\dfrac{1}{X}} \right)^4}{ + ^6}{C_5}{\left( X \right)^1}{\left( {\dfrac{1}{X}} \right)^5}{ + ^6}{C_6}{\left( {\dfrac{1}{X}} \right)^6}
Use the formula of combinations to expand each term
⇒(X+1X)6=6!(6−0)!0!(X)6+6!(6−1)!1!(X)5(1X)1+6!(6−2)!2!(X)4(1X)2+6!(6−3)!3!(X)3(1X)3\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = \dfrac{{6!}}{{(6 - 0)!0!}}{\left( X \right)^6} + \dfrac{{6!}}{{(6 - 1)!1!}}{\left( X \right)^5}{\left( {\dfrac{1}{X}} \right)^1} + \dfrac{{6!}}{{(6 - 2)!2!}}{\left( X \right)^4}{\left( {\dfrac{1}{X}} \right)^2} + \dfrac{{6!}}{{(6 - 3)!3!}}{\left( X \right)^3}{\left( {\dfrac{1}{X}} \right)^3}
+6!(6−4)!4!(X)2(1X)4+6!(6−5)!5!(X)1(1X)5+6!(6−6)!6!(1X)6+ \dfrac{{6!}}{{(6 - 4)!4!}}{\left( X \right)^2}{\left( {\dfrac{1}{X}} \right)^4} + \dfrac{{6!}}{{(6 - 5)!5!}}{\left( X \right)^1}{\left( {\dfrac{1}{X}} \right)^5} + \dfrac{{6!}}{{(6 - 6)!6!}}{\left( {\dfrac{1}{X}} \right)^6}
Solve the factorial subtraction
⇒(X+1X)6=6!6!0!(X)6+6!5!1!(X)5(1X)1+6!4!2!(X)4(1X)2+6!3!3!(X)3(1X)3\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = \dfrac{{6!}}{{6!0!}}{\left( X \right)^6} + \dfrac{{6!}}{{5!1!}}{\left( X \right)^5}{\left( {\dfrac{1}{X}} \right)^1} + \dfrac{{6!}}{{4!2!}}{\left( X \right)^4}{\left( {\dfrac{1}{X}} \right)^2} + \dfrac{{6!}}{{3!3!}}{\left( X \right)^3}{\left( {\dfrac{1}{X}} \right)^3}
+6!2!4!(X)2(1X)4+6!1!5!(X)1(1X)5+6!0!6!(1X)6+ \dfrac{{6!}}{{2!4!}}{\left( X \right)^2}{\left( {\dfrac{1}{X}} \right)^4} + \dfrac{{6!}}{{1!5!}}{\left( X \right)^1}{\left( {\dfrac{1}{X}} \right)^5} + \dfrac{{6!}}{{0!6!}}{\left( {\dfrac{1}{X}} \right)^6}
Put the value of 0!=10! = 1
⇒(X+1X)6=6!6!(X)6+6!5!(X)5(1X)1+6!4!2!(X)4(1X)2+6!3!3!(X)3(1X)3\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = \dfrac{{6!}}{{6!}}{\left( X \right)^6} + \dfrac{{6!}}{{5!}}{\left( X \right)^5}{\left( {\dfrac{1}{X}} \right)^1} + \dfrac{{6!}}{{4!2!}}{\left( X \right)^4}{\left( {\dfrac{1}{X}} \right)^2} + \dfrac{{6!}}{{3!3!}}{\left( X \right)^3}{\left( {\dfrac{1}{X}} \right)^3}
+6!2!4!(X)2(1X)4+6!5!(X)1(1X)5+6!6!(1X)6+ \dfrac{{6!}}{{2!4!}}{\left( X \right)^2}{\left( {\dfrac{1}{X}} \right)^4} + \dfrac{{6!}}{{5!}}{\left( X \right)^1}{\left( {\dfrac{1}{X}} \right)^5} + \dfrac{{6!}}{{6!}}{\left( {\dfrac{1}{X}} \right)^6}
Use the formula of factorial to open terms in numerator such that we can cancel as many terms as possible from the denominator
⇒(X+1X)6=(X)6+6×5!5!(X)5(1X)1+6×5×4!4!2!(X)4(1X)2+6×5×4×3!3!3!(X)3(1X)3\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = {\left( X \right)^6} + \dfrac{{6 \times 5!}}{{5!}}{\left( X \right)^5}{\left( {\dfrac{1}{X}} \right)^1} + \dfrac{{6 \times 5 \times 4!}}{{4!2!}}{\left( X \right)^4}{\left( {\dfrac{1}{X}} \right)^2} + \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!3!}}{\left( X \right)^3}{\left( {\dfrac{1}{X}} \right)^3}
+6×5×4!2!4!(X)2(1X)4+6×5!5!(X)1(1X)5+(1X)6+ \dfrac{{6 \times 5 \times 4!}}{{2!4!}}{\left( X \right)^2}{\left( {\dfrac{1}{X}} \right)^4} + \dfrac{{6 \times 5!}}{{5!}}{\left( X \right)^1}{\left( {\dfrac{1}{X}} \right)^5} + {\left( {\dfrac{1}{X}} \right)^6}
Cancel possible terms from denominator and numerator
⇒(X+1X)6=X6+6X51X+15X41X2+20X31X3+15X21X4+6X1X5+1X6\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = {X^6} + 6{X^5}\dfrac{1}{X} + 15{X^4}\dfrac{1}{{{X^2}}} + 20{X^3}\dfrac{1}{{{X^3}}} + 15{X^2}\dfrac{1}{{{X^4}}} + 6X\dfrac{1}{{{X^5}}} + \dfrac{1}{{{X^6}}}
Now we collect the powers of the same variable i.e. X
⇒(X+1X)6=X6+6X5−1+15X4−2+20X3−3+15X2−4+6X1−5+X0−6\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = {X^6} + 6{X^{5 - 1}} + 15{X^{4 - 2}} + 20{X^{3 - 3}} + 15{X^{2 - 4}} + 6{X^{1 - 5}} + {X^{0 - 6}}
⇒(X+1X)6=X6+6X4+15X2+20X0+15X−2+6X−4+X−6\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = {X^6} + 6{X^4} + 15{X^2} + 20{X^0} + 15{X^{ - 2}} + 6{X^{ - 4}} + {X^{ - 6}}
Put the term having power 0 equal to 1
⇒(X+1X)6=X6+6X4+15X2+20+15X−2+6X−4+X−6\Rightarrow {\left( {X + \dfrac{1}{X}} \right)^6} = {X^6} + 6{X^4} + 15{X^2} + 20 + 15{X^{ - 2}} + 6{X^{ - 4}} + {X^{ - 6}}
∴\therefore The expansion of (X+1X)6{\left( {X + \dfrac{1}{X}} \right)^6} using binomial theorem is given as X6+6X4+15X2+20+15X−2+6X−4+X−6{X^6} + 6{X^4} + 15{X^2} + 20 + 15{X^{ - 2}} + 6{X^{ - 4}} + {X^{ - 6}}

Note: Do not make mistake of collecting or adding the terms when the variables are in addition also, this is wrong, Keep in mind we add or combine the powers when we have multiplication of same variable i.e. ap.aq=ap+q{a^p}.{a^q} = {a^{p + q}} .