The sum of rational terms in the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ is.
(a) 31
(b) 41
(c) 51
(d) None of these
Answer
632.1k+ views
Hint: For solving this question first, we will expand the given term using the binomial expansion formulae and then find the rational terms. Then we will add them to get the correct answer.
Complete step-by-step answer:
Given:
We have to find the sum of rational terms in the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$.
In the expansion of the ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ , terms which are rational will not have fractional power, on the other hand, irrational terms will have fractional powers.
Now, we will use the following binomial expansion result:
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}\cdot y+{}^{n}{{C}_{2}}{{x}^{n-2}}\cdot {{y}^{2}}+{}^{n}{{C}_{3}}{{x}^{n-3}}\cdot {{y}^{3}}+{}^{n}{{C}_{4}}{{x}^{n-4}}\cdot {{y}^{4}}+{}^{n}{{C}_{5}}{{x}^{n-5}}\cdot {{y}^{5}}+....................+{}^{n}{{C}_{n}}{{y}^{n}}\]
Where, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
We can apply the above result to expand ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ . Then,
\[\begin{align}
& {{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}={}^{10}{{C}_{0}}{{\left( \sqrt{2} \right)}^{10}}+{}^{10}{{C}_{1}}{{\left( \sqrt{2} \right)}^{9}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{1}}+{}^{10}{{C}_{2}}{{\left( \sqrt{2} \right)}^{8}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{2}}+{}^{10}{{C}_{3}}{{\left( \sqrt{2} \right)}^{7}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{3}}+{}^{10}{{C}_{4}}{{\left( \sqrt{2} \right)}^{6}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{4}} \\
& +{}^{10}{{C}_{5}}{{\left( \sqrt{2} \right)}^{5}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{5}}\text{ }+{}^{10}{{C}_{6}}{{\left( \sqrt{2} \right)}^{4}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{6}}+{}^{10}{{C}_{7}}{{\left( \sqrt{2} \right)}^{3}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{7}}+{}^{10}{{C}_{8}}{{\left( \sqrt{2} \right)}^{2}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{8}}+{}^{10}{{C}_{9}}{{\left( \sqrt{2} \right)}^{1}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{9}}+{}^{10}{{C}_{10}}{{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{10}} \\
& \Rightarrow {{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}={}^{10}{{C}_{0}}{{\left( 2 \right)}^{5}}+{}^{10}{{C}_{1}}{{\left( 2 \right)}^{{}^{9}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{1}/{}_{5}}}+{}^{10}{{C}_{2}}{{\left( 2 \right)}^{4}}\cdot {{\left( 3 \right)}^{{}^{2}/{}_{5}}}+{}^{10}{{C}_{3}}{{\left( 2 \right)}^{{}^{7}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{3}/{}_{5}}}+{}^{10}{{C}_{4}}{{\left( 2 \right)}^{3}}\cdot {{\left( 3 \right)}^{{}^{4}/{}_{5}}}+{}^{10}{{C}_{5}}{{\left( 2 \right)}^{{}^{5}/{}_{2}}}\cdot {{\left( 3 \right)}^{1}} \\
& \text{ }+{}^{10}{{C}_{6}}{{\left( 2 \right)}^{2}}\cdot {{\left( 3 \right)}^{{}^{6}/{}_{5}}}+{}^{10}{{C}_{7}}{{\left( 2 \right)}^{{}^{3}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{7}/{}_{5}}}+{}^{10}{{C}_{8}}{{\left( 2 \right)}^{1}}\cdot {{\left( 3 \right)}^{{}^{8}/{}_{5}}}+{}^{10}{{C}_{9}}{{\left( 2 \right)}^{{}^{1}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{9}/{}_{5}}}+{}^{10}{{C}_{10}}{{\left( 3 \right)}^{2}} \\
\end{align}\]
Thus, in the above expression, we can see that there are only two rational terms in the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ . These rational terms are the first term and the last term of the expansion. Rational terms are \[{}^{10}{{C}_{0}}{{\left( 2 \right)}^{5}}=32\] and \[{}^{10}{{C}_{10}}{{\left( 3 \right)}^{2}}=9\] .
Now, adding the rational terms of the expansion. Then,
\[\begin{align}
& {}^{10}{{C}_{0}}{{\left( 2 \right)}^{5}}+{}^{10}{{C}_{10}}{{\left( 3 \right)}^{2}} \\
& \Rightarrow 32+9 \\
& \Rightarrow 41 \\
\end{align}\]
Thus, the sum of rational terms in the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ is 41.
Hence, option (b) is the correct option.
Note: Here, the student should write the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ correctly as per the binomial expansion formula without missing any term. Then, check each term clearly whether it is rational or irrational then solve for the correct answer without any calculation mistake.
Complete step-by-step answer:
Given:
We have to find the sum of rational terms in the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$.
In the expansion of the ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ , terms which are rational will not have fractional power, on the other hand, irrational terms will have fractional powers.
Now, we will use the following binomial expansion result:
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}\cdot y+{}^{n}{{C}_{2}}{{x}^{n-2}}\cdot {{y}^{2}}+{}^{n}{{C}_{3}}{{x}^{n-3}}\cdot {{y}^{3}}+{}^{n}{{C}_{4}}{{x}^{n-4}}\cdot {{y}^{4}}+{}^{n}{{C}_{5}}{{x}^{n-5}}\cdot {{y}^{5}}+....................+{}^{n}{{C}_{n}}{{y}^{n}}\]
Where, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
We can apply the above result to expand ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ . Then,
\[\begin{align}
& {{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}={}^{10}{{C}_{0}}{{\left( \sqrt{2} \right)}^{10}}+{}^{10}{{C}_{1}}{{\left( \sqrt{2} \right)}^{9}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{1}}+{}^{10}{{C}_{2}}{{\left( \sqrt{2} \right)}^{8}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{2}}+{}^{10}{{C}_{3}}{{\left( \sqrt{2} \right)}^{7}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{3}}+{}^{10}{{C}_{4}}{{\left( \sqrt{2} \right)}^{6}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{4}} \\
& +{}^{10}{{C}_{5}}{{\left( \sqrt{2} \right)}^{5}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{5}}\text{ }+{}^{10}{{C}_{6}}{{\left( \sqrt{2} \right)}^{4}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{6}}+{}^{10}{{C}_{7}}{{\left( \sqrt{2} \right)}^{3}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{7}}+{}^{10}{{C}_{8}}{{\left( \sqrt{2} \right)}^{2}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{8}}+{}^{10}{{C}_{9}}{{\left( \sqrt{2} \right)}^{1}}\cdot {{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{9}}+{}^{10}{{C}_{10}}{{\left( {{3}^{{}^{1}/{}_{5}}} \right)}^{10}} \\
& \Rightarrow {{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}={}^{10}{{C}_{0}}{{\left( 2 \right)}^{5}}+{}^{10}{{C}_{1}}{{\left( 2 \right)}^{{}^{9}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{1}/{}_{5}}}+{}^{10}{{C}_{2}}{{\left( 2 \right)}^{4}}\cdot {{\left( 3 \right)}^{{}^{2}/{}_{5}}}+{}^{10}{{C}_{3}}{{\left( 2 \right)}^{{}^{7}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{3}/{}_{5}}}+{}^{10}{{C}_{4}}{{\left( 2 \right)}^{3}}\cdot {{\left( 3 \right)}^{{}^{4}/{}_{5}}}+{}^{10}{{C}_{5}}{{\left( 2 \right)}^{{}^{5}/{}_{2}}}\cdot {{\left( 3 \right)}^{1}} \\
& \text{ }+{}^{10}{{C}_{6}}{{\left( 2 \right)}^{2}}\cdot {{\left( 3 \right)}^{{}^{6}/{}_{5}}}+{}^{10}{{C}_{7}}{{\left( 2 \right)}^{{}^{3}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{7}/{}_{5}}}+{}^{10}{{C}_{8}}{{\left( 2 \right)}^{1}}\cdot {{\left( 3 \right)}^{{}^{8}/{}_{5}}}+{}^{10}{{C}_{9}}{{\left( 2 \right)}^{{}^{1}/{}_{2}}}\cdot {{\left( 3 \right)}^{{}^{9}/{}_{5}}}+{}^{10}{{C}_{10}}{{\left( 3 \right)}^{2}} \\
\end{align}\]
Thus, in the above expression, we can see that there are only two rational terms in the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ . These rational terms are the first term and the last term of the expansion. Rational terms are \[{}^{10}{{C}_{0}}{{\left( 2 \right)}^{5}}=32\] and \[{}^{10}{{C}_{10}}{{\left( 3 \right)}^{2}}=9\] .
Now, adding the rational terms of the expansion. Then,
\[\begin{align}
& {}^{10}{{C}_{0}}{{\left( 2 \right)}^{5}}+{}^{10}{{C}_{10}}{{\left( 3 \right)}^{2}} \\
& \Rightarrow 32+9 \\
& \Rightarrow 41 \\
\end{align}\]
Thus, the sum of rational terms in the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ is 41.
Hence, option (b) is the correct option.
Note: Here, the student should write the expansion of ${{\left( \sqrt{2}+{{3}^{{}^{1}/{}_{5}}} \right)}^{10}}$ correctly as per the binomial expansion formula without missing any term. Then, check each term clearly whether it is rational or irrational then solve for the correct answer without any calculation mistake.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

