Answer
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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.
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