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Question: Find the \({4^{th}}\) term from the beginning and \({4^{th}}\) term from the end in the expansion of...

Find the 4th{4^{th}} term from the beginning and 4th{4^{th}} term from the end in the expansion of (x+2x)9{\left( {x + \dfrac{2}{x}} \right)^9}.
A. 652x2,5436x2652{x^2},\dfrac{{5436}}{{{x^2}}}
B. 672x3,5376x3672{x^3},\dfrac{{5376}}{{{x^3}}}
C. 672x4,536x2672{x^4},\dfrac{{536}}{{{x^2}}}
D. None of these

Explanation

Solution

We will find the value of 4th{4^{th}}term. So, we will break T4{T_4}into two parts such that T3+1{T_{3 + 1}}and then we will use a term or combination method to find the 4th term from the beginning and 4th term from the end. By using (x+a)n=∑r=0n  nCr xn−r ar.{\left( {x + a} \right)^n} = \sum\limits_{r = 0}^n {\,\,{n_{Cr}}\,{x^{n - r}}\,{a^r}.}

Complete step by step solution:
Solved by binomial theorem.
(x+2x)9{\left( {x + \dfrac{2}{x}} \right)^9} .
T4=T3+1=(x+2x)9{T_4} = {T_{3 + 1}} = {\left( {x + \dfrac{2}{x}} \right)^9}
r=3r = 3
T4= 9C3(x)9−3(2x)3{T_4} = {\,^9}{C_3}{\left( x \right)^{9 - 3}}{\left( {\dfrac{2}{x}} \right)^3}
T4= 9C3 x6.2×2×2x3{T_4} = {\,^9}{C_3}\,{x^6}.\dfrac{{2 \times 2 \times 2}}{{{x^3}}}
T4=9!6! 3!x6×8x3{T_4} = \dfrac{{9!}}{{6!\,3!}}{x^6} \times \dfrac{8}{{{x^3}}}
T4=9×8×7×6!3×2×1×6!x6×8x3{T_4} = \dfrac{{9 \times 8 \times 7 \times 6!}}{{3 \times 2 \times 1 \times 6!}}{x^6} \times \dfrac{8}{{{x^3}}}
T4=672 x3{T_4} = 672\,{x^3}
T7= 9C9−3 x3(2x)6{T_7} = {\,^9}{C_{9 - 3}}\,{x^3}{\left( {\dfrac{2}{x}} \right)^6}
T7= 9C6 x3(2x)6{T_7} = {\,^9}{C_6}\,{x^3}{\left( {\dfrac{2}{x}} \right)^6}
T7=9!6! 3! x3×2×2×2×2×2×2x6{T_7} = \dfrac{{9!}}{{6!\,3!}}\,{x^3} \times \dfrac{{2 \times 2 \times 2 \times 2 \times 2 \times 2}}{{{x^6}}}
T7=9×8×7×6!3×2×1  6!×64x6×x3{T_7} = \dfrac{{9 \times 8 \times 7 \times 6!}}{{3 \times 2 \times 1\,\,6!}} \times \dfrac{{64}}{{{x^6}}} \times {x^3}
T7=5376x3{T_7} = \dfrac{{5376}}{{{x^3}}}

Note: Students should carefully solve the value of factorial otherwise you will get the wrong answer and also break the value T4{T_4} into two parts for applying the binomial theorem.