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Question: The number of irrational terms in the expansion of \({{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{1...

The number of irrational terms in the expansion of (415+7110)45{{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{10}}} \right)}^{45}} is?
(a) 40
(b) 5
(c) 41
(d) None of these

Explanation

Solution

Use the fact that (x+y)n{{\left( x+y \right)}^{n}} has (n + 1) terms and try to find the number of rational terms in the expression (415+7110)45{{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{10}}} \right)}^{45}}. Use the formula for the general term of the expression (x+y)n{{\left( x+y \right)}^{n}} given as Tr+1=nCrxnryr{{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}} and simplify the general term of the expression (415+7110)45{{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{10}}} \right)}^{45}}. Check for which values of r we will get the rational terms. Subtract the number of rational terms from the total number of terms to get the number of irrational terms.

Complete step by step answer:
Here we have been provided with the expression (415+7110)45{{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{10}}} \right)}^{45}} and we are asked number of irrational terms in its expansion. Here we will find the number of rational terms and then subtract it from the total number of terms to get the answer. Let us assume the expression as E, so we have,
E=(415+7110)45\Rightarrow E={{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{10}}} \right)}^{45}}
We know that number of terms in the binomial expression (x+y)n{{\left( x+y \right)}^{n}} is (n + 1), so the total number of terms in (415+7110)45{{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{10}}} \right)}^{45}} will be 46. Now, the general term of the expression (x+y)n{{\left( x+y \right)}^{n}} is given as Tr+1=nCrxnryr{{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}, so replacing x with 415{{4}^{\dfrac{1}{5}}} and y with 7110{{7}^{\dfrac{1}{10}}} we get the general term of (415+7110)45{{\left( {{4}^{\dfrac{1}{5}}}+{{7}^{\dfrac{1}{10}}} \right)}^{45}} as: -
Tr+1=45Cr(415)45r(7110)r Tr+1=45Cr(445r5)(7r10) Tr+1=45Cr(49r5)(7r10) \begin{aligned} & \Rightarrow {{T}_{r+1}}={}^{45}{{C}_{r}}{{\left( {{4}^{\dfrac{1}{5}}} \right)}^{45-r}}{{\left( {{7}^{\dfrac{1}{10}}} \right)}^{r}} \\\ & \Rightarrow {{T}_{r+1}}={}^{45}{{C}_{r}}\left( {{4}^{\dfrac{45-r}{5}}} \right)\left( {{7}^{\dfrac{r}{10}}} \right) \\\ & \Rightarrow {{T}_{r+1}}={}^{45}{{C}_{r}}\left( {{4}^{9-\dfrac{r}{5}}} \right)\left( {{7}^{\dfrac{r}{10}}} \right) \\\ \end{aligned}
Using the formula of exponents given as amn=aman{{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}} we get,
Tr+1=45Cr(49)(7r104r5)\Rightarrow {{T}_{r+1}}={}^{45}{{C}_{r}}\left( {{4}^{9}} \right)\left( \dfrac{{{7}^{\dfrac{r}{10}}}}{{{4}^{\dfrac{r}{5}}}} \right)
Now, on observing the above relation of general terms we can say that the terms will be rational if and only if the value of r will be a multiple of 10, so the values of r can be 0, 1, 2, 3 and 4. We can see that there are 5 values of r so there will be five rational terms.
\Rightarrow Number of irrational terms = 46 – 5
\Rightarrow Number of irrational terms = 41

So, the correct answer is “Option c”.

Note: As you can see that there are 41 irrational terms in the expansion of the given expression so there will be 41 such values of r. This is the reason we haven’t taken those values of r as found the rational terms first which were only 5 in number. You must understand the condition so that it may take less time to solve the questions.