Question

Write down and simplify :
The $(r+1{)}^{\text{th}}$ term of $(1+x{)}^{\frac{11}{3}}$

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,

$T_{r+1}=\frac{\frac{11}{3}(\frac{11}{3}-1)(\frac{11}{3}-2)(\frac{11}{3}-3)(\frac{11}{3}-4)..........(\frac{11}{3}-r+1)}{r!}(x{)}^r$

$=\frac{\mathrm{11.8.5.2.}(-1).........(14-3r)}{3^r.r!}(x{)}^r$

$=(-1{)}^{r-1}\frac{\mathrm{11.8.5.2.1.4......}(3r-14)}{3^r.r!}{x}^r$