Question

If \[{{\left( {{5}^{\dfrac{1}{2}}}+{{7}^{\dfrac{1}{6}}} \right)}^{642}}\] contains ‘n’ integral terms then n is.
A. \[108\]
B. \[106\]
C. \[107\]
D. \[109\]

Answer 1

Hint: In the given question, we have been asked to find out the number of integral term of the given expression. In order to find out the numbers of integral terms, we will use the method of expansion and as we know that the general term of the expansion \[{{\left( a+b \right)}^{n}}\] is \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\] . Applying this we will substitute the value of \[a={{5}^{\dfrac{1}{2}}},\ b={{7}^{\dfrac{1}{6}}}\ and\ n=642\] and expand the given expression. Later as we know that the term to be integral it should be an integer. Thus in this way we will find out the number of integral terms.
Formula used:
The general term of the expansion \[{{\left( a+b \right)}^{n}}\] is given by,
 \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]

Complete step by step solution:
We have given that,
 \[\Rightarrow {{\left( {{5}^{\dfrac{1}{2}}}+{{7}^{\dfrac{1}{6}}} \right)}^{642}}\]
As we know that,
The general term of the expansion \[{{\left( a+b \right)}^{n}}\] is;
 \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]
Therefore,
For \[{{\left( {{5}^{\dfrac{1}{2}}}+{{7}^{\dfrac{1}{6}}} \right)}^{642}}\]
Here, \[a={{5}^{\dfrac{1}{2}}},\ b={{7}^{\dfrac{1}{6}}}\ and\ n=642\]
Thus, the general term of the expansion \[{{\left( {{5}^{\dfrac{1}{2}}}+{{7}^{\dfrac{1}{6}}} \right)}^{642}}\] is,
 \[{{T}_{r+1}}={}^{642}{{C}_{r}}{{\left( {{5}^{\dfrac{1}{2}}} \right)}^{642-r}}{{\left( {{7}^{\dfrac{1}{6}}} \right)}^{r}}\]
Simplifying the above expression, we will get
 \[{{T}_{r+1}}={}^{642}{{C}_{r}}\cdot {{5}^{\dfrac{642-r}{2}}}\cdot {{7}^{\dfrac{r}{6}}}\]
Therefore,
For the term to be integral \[\dfrac{r}{6}\] should be an integer.
Thus the possible values of ‘r’ are;
The values of ‘r’ includes all the multiples of 6 up to the number 642.
 \[\Rightarrow r=0,6,12,18,24,30,.......642\]
Also,
We can observe that for the above values of ‘r’ the term \[\dfrac{642-r}{2}\] is also integral.
Therefore,
Number of integral terms are = \[r+1=\dfrac{642}{6}+1=107+1=108\]
Hence there are a total of 108 integral terms.
Thus, the option (a) is the correct answer.
So, the correct answer is “Option a”.

Note: While solving these types of question where we have been asked to find out the number of integral terms, students need to keep in mind that in a binomial expansion for a term to be an integral it should be an integer that is it should have fractional powers. This means that the terms should not have square roots, cube roots or fourths root.