Question

Find the 4th term from the beginning and 4th term from the end in the expansion of ${\left(x+\frac{2}{x}\right)}^9$.

Answer 1

Let T_r+1 be the 4th term from the end.
Then, T_r+1 is (10 − 4 + 1)th, i.e., 7th, term from the beginning.

$$\begin{gathered} \therefore T_7=T_{6+1} \\ ={}^{9}C_6\left(x^{9-6}\right){\left(\frac{2}{x}\right)}^6 \\ =\frac{9\times 8\times 7}{3\times 2}\left(x^3\right)\left(\frac{64}{x^6}\right) \\ =\frac{5376}{x^3} \end{gathered}$$

4th term from the beginning = $T_4=T_{3+1}$
$$\begin{gathered} \therefore T_4={}^{9}C_3\left(x^{9-3}\right){\left(\frac{2}{x}\right)}^3 \\ =\frac{9\times 8\times 7}{3\times 2}\left(x^6\right)\left(\frac{8}{x^3}\right) \\ =672x^3 \end{gathered}$$