Question

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

Answer 1

Hint: We will find the value of ${4^{th}}$term. So, we will break ${T_4}$into two parts such that ${T_{3 + 1}}$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 ${\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.
${\left( {x + \dfrac{2}{x}} \right)^9}$ .
${T_4} = {T_{3 + 1}} = {\left( {x + \dfrac{2}{x}} \right)^9}$
$r = 3$
${T_4} = {\,^9}{C_3}{\left( x \right)^{9 - 3}}{\left( {\dfrac{2}{x}} \right)^3}$
${T_4} = {\,^9}{C_3}\,{x^6}.\dfrac{{2 \times 2 \times 2}}{{{x^3}}}$
${T_4} = \dfrac{{9!}}{{6!\,3!}}{x^6} \times \dfrac{8}{{{x^3}}}$
${T_4} = \dfrac{{9 \times 8 \times 7 \times 6!}}{{3 \times 2 \times 1 \times 6!}}{x^6} \times \dfrac{8}{{{x^3}}}$
${T_4} = 672\,{x^3}$
 ${T_7} = {\,^9}{C_{9 - 3}}\,{x^3}{\left( {\dfrac{2}{x}} \right)^6}$
${T_7} = {\,^9}{C_6}\,{x^3}{\left( {\dfrac{2}{x}} \right)^6}$
${T_7} = \dfrac{{9!}}{{6!\,3!}}\,{x^3} \times \dfrac{{2 \times 2 \times 2 \times 2 \times 2 \times 2}}{{{x^6}}}$
${T_7} = \dfrac{{9 \times 8 \times 7 \times 6!}}{{3 \times 2 \times 1\,\,6!}} \times \dfrac{{64}}{{{x^6}}} \times {x^3}$
${T_7} = \dfrac{{5376}}{{{x^3}}}$


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