Question

Write down and simplify :
The ${14}^{\text{th}}$ term of $(2^{10}-2^{7}x{)}^{\frac{13}{2}}$

Answer 1

The general term i.e. $(r+1)th$ term in the expansion of $(1+x{)}^n$ is given by

$T_{r+1}=\frac{n(n-1)(n-2)....(n-r+1)}{r!}{x}^r$

Now,

$(2^{10}-2^{7}x{)}^{\frac{13}{2}}=2^{65}{(1-\frac{2^7}{2^{10}}x)}^{\frac{13}{2}}=2^{65}{(1-\frac{x}{8})}^{\frac{13}{2}}$

For $T_{14},$ $r=13$

$\therefore T_{14}=2^{65}\left[\frac{\frac{13}{2}(\frac{13}{2}-1)(\frac{13}{2}-2)(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-5)(\frac{13}{2}-6)(\frac{13}{2}-7)(\frac{13}{2}-8)(\frac{13}{2}-9)(\frac{13}{2}-10)(\frac{13}{2}-11)(\frac{13}{2}-12)}{13!}{(-\frac{x}{8})}^{13}\right]$

$=2^{65}\left[\frac{\mathrm{13.11.9.7.5.3.1.}(-1).(-3).(-5).(-7).(-9).(-11)}{2^{13}.13!}\right(-1{)}^{13}.\frac{1}{8^{13}}.x^{13}]$

$=-2^{65}[\frac{\mathrm{1.3.5.7.9.11}}{2^{13}.2.4.6.8.10.12}.\frac{1}{2^{39}}.x^{13}]$

$=-\frac{2^{65}}{2^{62}}.\left(\mathrm{7.3.11}\right)x^{13}$

$=-\mathrm{3.7.8.11.}{x}^{13}$

$=-1848x^{13}$