Question

Find the $(r+1{)}^{\text{th}}$ term of the following expansion :
$(1+3x{)}^{\frac{1}{3}}$

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,

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

$=\frac{1.(-2)(-5)(-8)........(4-3r)}{3^r.r!}(3x{)}^r$

$=\frac{1.(-2)(-5)(-8)........(4-3r)}{3^r.r!}{3}^r.x^r$

$=(-1{)}^{r-1}\frac{\mathrm{1.2.5.8......}(3r-4)}{r!}{x}^r$