Question

Find the $7^{th}$ term of $(1-2x{)}^9$ when $|x|<\frac{1}{2}$

Answer 1

$(1-2x{)}^9$ is of the form $(1+x{)}^n$

$(1+p{)}^n{=}^n{C}_0(1\left){}^n\right(p{)}^0{+}^n{C}_1\left(1{)}^{n-1}\right(p{)}^1+.......{.}^n{C}_n\left(1{)}^0\right(p{)}^1$

$p=-2x$

$n=9$

$(1-2x{)}^9{=}^n{C}_0(1{)}^9(-2x{)}^0{+}^9{C}_1(1{)}^8(-2x{)}^1+.....$

$r^{th}term{=}^n{C}_{(r-1)}\left(1{)}^{(n-r+1)}\right(-2C{)}^{(r-1)}$

$7^{th}term{=}^n{C}_6\left(1{)}^{(9-6+1)}\right(-2x{)}^6$

${=}^n{C}_6(2{)}^6{x}^6$

$7^{th}term=2^6{}^n{C}_r{x}^6$