Question

find the 11th term in the expansion of (x+3/x)²⁰​

Answer 1

Answer:

(2x−x21)25

Total terms = 26

11th term from last = 16th term from start.

⇒n=15

tn+1=mCn.am−nbn in (a+b)m

∴t16=26C15(2n)11(x2−1)15

=−26C15×211×x30x11

=−26C15×211×x−19

Answer 2

Step-by-step explanation:

Middle term = 11th term

$\sf \: = T_{ \: r+1} \\ \sf = { {}^{n}C_r ...x {}^{n - r} {a}^{r} } \\ \sf \: = \: {}^{20}C_r \bigg(x {}^{2} \bigg) {}^{20 - r \bigg({ \dfrac{3}{r} \bigg) }^{r} }$

put r = 10

$\sf{T_{10+1}}$$\sf \: = \: {}^{20}C_{10} \bigg(x {}^{2} \bigg) {}^{10 \bigg({{\dfrac{3}{x} \bigg) }^{} {}^{10} } }$

$\sf= \: {}^{20}C_{10} \: \: x {}^{20} {}^{10 {{\dfrac{3 {}^{10} }{x {}^{10} } }}}$

$\color{lime}\sf \: T_{10+1} = \: {}^{20}C_{10} \: \: x {}^{10} {3}^{10 }$