Question

The number of integral terms in the expansion of ${{\left( 2\sqrt{5}+\sqrt[6]{7} \right)}^{642}}$ are:
A. 105
B. 107
C. 321
D. 108

Answer 1

Binomial Expansion:
${{\left( a+b \right)}^{n}}={{C}_{0}}{{a}^{n}}+{{C}_{1}}{{a}^{n-1}}b+{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+\ldots +{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+\ldots +{{C}_{n-1}}a{{b}^{n-1}}+{{C}_{n}}{{b}^{n}}$ , where ${{C}_{0}},{{C}_{1}},...,{{C}_{n}}$ are the Binomial Coefficients defined as ${{C}_{r}}{{=}^{n}}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ .
The total number of terms in the expansion is $n+1$ .
The rth term in the expansion is ${{T}_{r}}={{C}_{r}}{{a}^{n-r}}{{b}^{r}}$ .
The value of $^{n}{{C}_{r}}$ is always an integer.
For the terms of the expansion to be an integer, the powers $x-r$ and $r$ should also be an integer because 5 and 7 are power free numbers.

Complete step-by-step answer:
Using the Binomial Expansion, the rth term of ${{T}_{r}}$ in the expansion of ${{\left( 2\sqrt{5}+\sqrt[6]{7} \right)}^{642}}$ will be given as:
${{T}_{r}}={{C}_{r}}{{\left( 2\sqrt{5} \right)}^{642-r}}{{\left( \sqrt[6]{7} \right)}^{r}}$
Since, $\sqrt{5}={{5}^{\dfrac{1}{2}}}$ and $\sqrt[6]{7}={{7}^{\dfrac{1}{6}}}$ , the above expression for ${{T}_{r}}$ can be written as:
${{T}_{r}}={{C}_{r}}{{\left( 2 \right)}^{642-r}}{{\left( 5 \right)}^{\dfrac{642-r}{2}}}{{\left( 7 \right)}^{\dfrac{r}{6}}}$
⇒ ${{T}_{r}}={{C}_{r}}{{\left( 2 \right)}^{642-r}}{{\left( 5 \right)}^{321-\dfrac{r}{2}}}{{\left( 7 \right)}^{\dfrac{r}{6}}}$
For the term to be an integer, we must have $0\le r\le 642$ and r should be a multiple of 6.
The required values of r are, therefore, the terms of the following AP:
$0,6,12,18,...,642$
i.e. $6\times 0,6\times 1,6\times 2,6\times 3,...,6\times 107$ .
∴ The total number of possible values of r are 108.
The correct answer option is D. 108.

Note: There are n integers from 1 to n, including both 1 and n.
The value of $^{n}{{C}_{r}}$ is always an integer.
${{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$, is not an integer unless $a=1$ or $x=0$ .
${{a}^{\dfrac{p}{q}}}=\sqrt[q]{{{a}^{p}}}$ .