Question

The number of integral terms in the expansion of $(2\sqrt{5}+\sqrt[6]{67}{)}^{642}$

• A. $105$
• B. $107$
• C. $321$
• D. $108$

Answer 1

The correct option is D $108$

We have, $(2\sqrt{5}+6\sqrt{7}{)}^{642}$

Now, General term of this binomial will be

$T_{r+1}={}^{642}{C}_r{\left(2\sqrt{5}\right)}^{642-r}(6\sqrt{7}{)}^r$

The general term will be an integer if ${}^{642}{C}_r$

is an integer and $(2\sqrt{5}{)}^{642-r}$ is an

integer and $(6\sqrt{7}{)}^r$ is an integer.

Now,

${}^{642}{C}_r$ will always to be a positive integer.

sice, ${}^{642}{C}_r$ denotes number of ways of selecting $r$ things out of $78$ things, it cannot be a fraction.

If $(2\sqrt{5}{)}^{642-r}$ is an integer.

Then, $642-r$ is an integer.

Hence, $r=0,2,4,6,\mathrm{8............640},642----\left(1\right)$

Also $7$ is an integer if $r/6$ is an integer.

$r=0,16,12,\mathrm{18.....}\left(2\right)$

from equation $\left(1\right)$ and $\left(2\right)$ to,

$r=0,6,12,18,\mathrm{24.....642}$

then solve by $A.P$ formula-

$a=0,l=642$

$n=?,d=6-0=6$

$T_n=l=a+(n-1)d$ ( using formula)

$642=0+(n-1)\times 6$

$n-1=\frac{642}{6}$

$n-1=107$

$n=108$

therefore, total $108$ terms

Hence this is the answer.