Find the 4 th term from the beginning and 4 th term from the end in the expansion of $\left(x+\frac{2}{x}\right)^{9}$.
Let Tr+1 be the 4th term from the end.
Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, term from the beginning.
$\therefore T_{7}=T_{6+1}$
$={ }^{9} C_{6}\left(x^{9-6}\right)\left(\frac{2}{x}\right)^{6}$
$=\frac{9 \times 8 \times 7}{3 \times 2}\left(x^{3}\right)\left(\frac{64}{x^{6}}\right)$
$=\frac{5376}{x^{3}}$
4th term from the beginning $=T_{4}=T_{3+1}$
$\therefore T_{4}={ }^{9} C_{3}\left(x^{9-3}\right)\left(\frac{2}{x}\right)^{3}$
$=\frac{9 \times 8 \times 7}{3 \times 2}\left(x^{6}\right)\left(\frac{8}{x^{3}}\right)$
$=672 x^{3}$
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