Question

The number of integral terms in the expansion of ${{\left( {{5}^{\dfrac{1}{2}}}+{{7}^{\dfrac{1}{6}}} \right)}^{642}}$ is:
(a) 108
(b) 106
(c) 103
(d) 109

Answer 1

Hint:First we will look at the number of terms in the expansion of ${{\left( x+1 \right)}^{n}}$ and then with the help of that we will write the total number of terms and then we will find the LCM of 2,6 and then all the terms which has power in multiple of 6 will be integer and then we will count them to get the final answer.

Complete step-by-step answer:
We the binomial expansion of ${{\left( x+1 \right)}^{n}}$ is:
${}^{n}{{c}_{0}}+{}^{n}{{c}_{1}}x+.....+{}^{n}{{c}_{n}}{{x}^{n}}$
We can see that it has total n+1 terms.
Now the LCM of 2 and 6 is 6.
We will also check if 642 is divisible by 6 or not.
$\dfrac{642}{6}=107$
Hence we will look at all the terms with power 0,6,12,18, …. 642.
Because then it will give all terms having integer values in the expansion ${{\left( {{5}^{\dfrac{1}{2}}}+{{7}^{\dfrac{1}{6}}} \right)}^{642}}$
When we divide 642 by 6 we get 107 and the first term is also integer.
Therefore the total integral terms will be 107+1 = 108.
Hence the correct answer is option (a).

Note: Another method to solve this question is to expand the given expression ${{\left( {{5}^{\dfrac{1}{2}}}+{{7}^{\dfrac{1}{6}}} \right)}^{642}}$by using the formula mentioned and then counting all the integral terms or one can also try to find out the pattern that after how much terms we get the integral term and with the help of that we can find the total integral terms, but it will be time consuming.