Question

In the expansion of $(7^{1/3}+{11}^{1/9}{)}^{6561}$ prove that three will be only $730$ term which are free from radicals

Answer 1

$T_{r+1}={}^{6561}{C}_r\left(7^{1/3}{)}^{6561-r}\right({11}^{1/9}{)}^r$
$T_{r+1}={}^{6561}{C}_r\left(7^{1/3}{)}^{6561-r}\right({11}^{1/9}{)}^r$
$={}^{6561}{C}_r{7}^{2187}\cdot 7^{-r/3}\cdot {11}^{r/9},0\le r\le 6561$
Now $T_{r+1}$ will be integral if r/3 and r/9 both are integers in the span $0\le r\le 6561$. It will be possible if r is multiple of 9. In other words, we have to find the numbers of those numbers lying between 0 to 6561 which are multiple of 9 i.e. those which form an A.P. of common difference is 9.
0,9,18,27, ....,6561
Note that 6561 is divisible by 9. If it were not divisible by 9 we would have chosen just the preceding number which is divisible by 9.
$T_n=6561=a+(n-1)d=0+(n-1)9$
$\therefore n=1+\frac{6561}{9}=1+729=730$
Hence there will be 730 terms will be free from radicals .