Find the number of rational terms present in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
(a) 6
(b) 7
(c) 3
(d) 8
Answer
Verified623.7k+ views
Hint:We start solving the problem by assuming $a=1+\sqrt{2}$ and $b=\sqrt[3]{3}$. We then recall the expansion of ${{\left( a+b \right)}^{n}}$ and use it for the given expansion. We then substitute$b=\sqrt[3]{3}$ in the obtained result and take out the terms that were not multiplied by the irrational terms. We then substitute $a=1+\sqrt{2}$ in the terms that were multiplied by the rational terms and expand them using the expansion ${{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$. We then separate the rational terms and count them to get the desired result.
Complete step by step answer:
According to the problem, we need to find the total number of rational terms present in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
Let us rewrite the terms the terms present in the given expansion as ${{\left( \left( 1+\sqrt{2} \right)+\sqrt[3]{3} \right)}^{6}}$. Let us assume $a=1+\sqrt{2}$ and $b=\sqrt[3]{3}$.
We know that the binomial expansion of ${{\left( a+b \right)}^{n}}$ is defined as ${}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+......+{}^{n}{{C}_{n}}{{b}^{n}}$.
So, we get ${{\left( a+b \right)}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{1}}{{a}^{6-1}}{{b}^{1}}+{}^{6}{{C}_{2}}{{a}^{6-2}}{{b}^{2}}+{}^{6}{{C}_{3}}{{a}^{6-3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{6-4}}{{b}^{4}}+{}^{6}{{C}_{5}}{{a}^{6-5}}{{b}^{1}}+{}^{6}{{C}_{6}}{{a}^{6-6}}{{b}^{6}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{6!}{0!6!}{{a}^{6}}+\dfrac{6!}{1!\left( 5 \right)!}{{a}^{6-1}}{{b}^{1}}+\dfrac{6!}{2!4!}{{a}^{6-2}}{{b}^{2}}+\dfrac{6!}{3!3!}{{a}^{6-3}}{{b}^{3}}+\dfrac{6!}{4!2!}{{a}^{6-4}}{{b}^{4}}+\dfrac{6!}{5!1!}{{a}^{6-5}}{{b}^{1}}+\dfrac{6!}{6!0!}{{a}^{6-6}}{{b}^{6}}\].
We know that $0!=1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{1}{1}{{a}^{6}}+\dfrac{6}{1}{{a}^{5}}{{b}^{1}}+\dfrac{6\times 5}{2\times 1}{{a}^{4}}{{b}^{2}}+\dfrac{6\times 5\times 4}{3\times 2\times 1}{{a}^{3}}{{b}^{3}}+\dfrac{6\times 5}{2\times 1}{{a}^{2}}{{b}^{4}}+\dfrac{6}{1}{{a}^{1}}{{b}^{1}}+\dfrac{1}{1}{{a}^{0}}{{b}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{b}^{1}}+15{{a}^{4}}{{b}^{2}}+20{{a}^{3}}{{b}^{3}}+15{{a}^{2}}{{b}^{4}}+6a{{b}^{5}}+{{b}^{6}}\].
Now, let us substitute $b=\sqrt[3]{3}$.
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{\left( \sqrt[3]{3} \right)}^{1}}+15{{a}^{4}}{{\left( \sqrt[3]{3} \right)}^{2}}+20{{a}^{3}}{{\left( \sqrt[3]{3} \right)}^{3}}+15{{a}^{2}}{{\left( \sqrt[3]{3} \right)}^{4}}+6a{{\left( \sqrt[3]{3} \right)}^{5}}+{{\left( \sqrt[3]{3} \right)}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}\left( \sqrt[3]{3} \right)+15{{a}^{4}}\left( \sqrt[3]{9} \right)+20{{a}^{3}}\left( 3 \right)+15{{a}^{2}}\left( 3\sqrt[3]{3} \right)+6a\left( 3\sqrt[3]{9} \right)+9\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6\sqrt[3]{3}{{a}^{5}}+15\sqrt[3]{9}{{a}^{4}}+60{{a}^{3}}+45\sqrt[3]{3}{{a}^{2}}+18\sqrt[3]{9}a+9\]---(1).
We can see that $\sqrt[3]{3}$ and $\sqrt[3]{9}$ are irrational and even if we multiply them with integer, we get irrational only. Since $\sqrt[3]{3}$, $\sqrt[3]{9}$ have different square root comparing with $\sqrt{2}$, we get irrational when they multiply with each other.
So, we now take the terms that are not multiplying with $\sqrt[3]{3}$ and $\sqrt[3]{9}$.
So, we get rational terms only in ${{a}^{6}}$, $60{{a}^{3}}$ and 9(which is already a rational number) ---(2).
Now, we expand ${{a}^{6}}={{\left( 1+\sqrt{2} \right)}^{6}}$. We know that ${{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$.
So, we get ${{a}^{6}}=1+{}^{6}{{C}_{1}}\left( \sqrt{2} \right)+{}^{6}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{6}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}+{}^{6}{{C}_{4}}{{\left( \sqrt{2} \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( \sqrt{2} \right)}^{5}}+{}^{6}{{C}_{6}}{{\left( \sqrt{2} \right)}^{6}}$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6!}{1!5!}\left( \sqrt{2} \right)+\dfrac{6!}{2!4!}\left( 2 \right)+\dfrac{6!}{3!3!}\left( 2\sqrt{2} \right)+\dfrac{6!}{4!2!}\left( 4 \right)+\dfrac{6!}{5!1!}\left( 4\sqrt{2} \right)+\dfrac{6!}{6!0!}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6}{1}\left( \sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 2 \right)+\dfrac{6\times 5\times 4}{3\times 2\times 1}\left( 2\sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 4 \right)+\dfrac{6}{1}\left( 4\sqrt{2} \right)+\dfrac{1}{1}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+6\sqrt{2}+30+40\sqrt{2}+60+24\sqrt{2}+8$.
We have got the terms 1, 30, 60 and 8 as rational which are 4 ---(3).
Now, let us expand ${{a}^{3}}={{\left( 1+\sqrt{2} \right)}^{3}}$.
So, we get ${{a}^{3}}=1+{}^{3}{{C}_{1}}\left( \sqrt{2} \right)+{}^{3}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}$.
$\Rightarrow {{a}^{3}}=1+\dfrac{3!}{1!2!}\left( \sqrt{2} \right)+\dfrac{3!}{2!1!}\left( 2 \right)+\dfrac{3!}{3!0!}\left( 2\sqrt{2} \right)$.
\[\Rightarrow {{a}^{3}}=1+\dfrac{3}{1}\left( \sqrt{2} \right)+\dfrac{3}{1}\left( 2 \right)+\dfrac{1}{1}\left( 2\sqrt{2} \right)\].
$\Rightarrow {{a}^{3}}=1+3\sqrt{2}+6+2\sqrt{2}$..
We have got the terms 1 and 6 as rational which are 2 ---(4).
From equations (2), (3), (4) we have got a total of $\left( 1+4+2 \right)=7$ rational terms.
∴ We have got a total 7 rational terms in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
The correct option for the given problem is (b).
Note:
We get only one rational if we add all the rational numbers we just obtained. Since they were asked about the number of rational terms present in it, we have taken them separately without adding. We can also take $a=1$ and $b=\sqrt{2}+\sqrt[3]{3}$ for solving this problem in which we just need to expand the powers of ‘b’ only to get the result. Similarly, we can expect problems to find the sum of the rational terms present in this expansion.
Complete step by step answer:
According to the problem, we need to find the total number of rational terms present in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
Let us rewrite the terms the terms present in the given expansion as ${{\left( \left( 1+\sqrt{2} \right)+\sqrt[3]{3} \right)}^{6}}$. Let us assume $a=1+\sqrt{2}$ and $b=\sqrt[3]{3}$.
We know that the binomial expansion of ${{\left( a+b \right)}^{n}}$ is defined as ${}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+......+{}^{n}{{C}_{n}}{{b}^{n}}$.
So, we get ${{\left( a+b \right)}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{1}}{{a}^{6-1}}{{b}^{1}}+{}^{6}{{C}_{2}}{{a}^{6-2}}{{b}^{2}}+{}^{6}{{C}_{3}}{{a}^{6-3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{6-4}}{{b}^{4}}+{}^{6}{{C}_{5}}{{a}^{6-5}}{{b}^{1}}+{}^{6}{{C}_{6}}{{a}^{6-6}}{{b}^{6}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{6!}{0!6!}{{a}^{6}}+\dfrac{6!}{1!\left( 5 \right)!}{{a}^{6-1}}{{b}^{1}}+\dfrac{6!}{2!4!}{{a}^{6-2}}{{b}^{2}}+\dfrac{6!}{3!3!}{{a}^{6-3}}{{b}^{3}}+\dfrac{6!}{4!2!}{{a}^{6-4}}{{b}^{4}}+\dfrac{6!}{5!1!}{{a}^{6-5}}{{b}^{1}}+\dfrac{6!}{6!0!}{{a}^{6-6}}{{b}^{6}}\].
We know that $0!=1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{1}{1}{{a}^{6}}+\dfrac{6}{1}{{a}^{5}}{{b}^{1}}+\dfrac{6\times 5}{2\times 1}{{a}^{4}}{{b}^{2}}+\dfrac{6\times 5\times 4}{3\times 2\times 1}{{a}^{3}}{{b}^{3}}+\dfrac{6\times 5}{2\times 1}{{a}^{2}}{{b}^{4}}+\dfrac{6}{1}{{a}^{1}}{{b}^{1}}+\dfrac{1}{1}{{a}^{0}}{{b}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{b}^{1}}+15{{a}^{4}}{{b}^{2}}+20{{a}^{3}}{{b}^{3}}+15{{a}^{2}}{{b}^{4}}+6a{{b}^{5}}+{{b}^{6}}\].
Now, let us substitute $b=\sqrt[3]{3}$.
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{\left( \sqrt[3]{3} \right)}^{1}}+15{{a}^{4}}{{\left( \sqrt[3]{3} \right)}^{2}}+20{{a}^{3}}{{\left( \sqrt[3]{3} \right)}^{3}}+15{{a}^{2}}{{\left( \sqrt[3]{3} \right)}^{4}}+6a{{\left( \sqrt[3]{3} \right)}^{5}}+{{\left( \sqrt[3]{3} \right)}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}\left( \sqrt[3]{3} \right)+15{{a}^{4}}\left( \sqrt[3]{9} \right)+20{{a}^{3}}\left( 3 \right)+15{{a}^{2}}\left( 3\sqrt[3]{3} \right)+6a\left( 3\sqrt[3]{9} \right)+9\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6\sqrt[3]{3}{{a}^{5}}+15\sqrt[3]{9}{{a}^{4}}+60{{a}^{3}}+45\sqrt[3]{3}{{a}^{2}}+18\sqrt[3]{9}a+9\]---(1).
We can see that $\sqrt[3]{3}$ and $\sqrt[3]{9}$ are irrational and even if we multiply them with integer, we get irrational only. Since $\sqrt[3]{3}$, $\sqrt[3]{9}$ have different square root comparing with $\sqrt{2}$, we get irrational when they multiply with each other.
So, we now take the terms that are not multiplying with $\sqrt[3]{3}$ and $\sqrt[3]{9}$.
So, we get rational terms only in ${{a}^{6}}$, $60{{a}^{3}}$ and 9(which is already a rational number) ---(2).
Now, we expand ${{a}^{6}}={{\left( 1+\sqrt{2} \right)}^{6}}$. We know that ${{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$.
So, we get ${{a}^{6}}=1+{}^{6}{{C}_{1}}\left( \sqrt{2} \right)+{}^{6}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{6}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}+{}^{6}{{C}_{4}}{{\left( \sqrt{2} \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( \sqrt{2} \right)}^{5}}+{}^{6}{{C}_{6}}{{\left( \sqrt{2} \right)}^{6}}$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6!}{1!5!}\left( \sqrt{2} \right)+\dfrac{6!}{2!4!}\left( 2 \right)+\dfrac{6!}{3!3!}\left( 2\sqrt{2} \right)+\dfrac{6!}{4!2!}\left( 4 \right)+\dfrac{6!}{5!1!}\left( 4\sqrt{2} \right)+\dfrac{6!}{6!0!}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6}{1}\left( \sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 2 \right)+\dfrac{6\times 5\times 4}{3\times 2\times 1}\left( 2\sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 4 \right)+\dfrac{6}{1}\left( 4\sqrt{2} \right)+\dfrac{1}{1}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+6\sqrt{2}+30+40\sqrt{2}+60+24\sqrt{2}+8$.
We have got the terms 1, 30, 60 and 8 as rational which are 4 ---(3).
Now, let us expand ${{a}^{3}}={{\left( 1+\sqrt{2} \right)}^{3}}$.
So, we get ${{a}^{3}}=1+{}^{3}{{C}_{1}}\left( \sqrt{2} \right)+{}^{3}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}$.
$\Rightarrow {{a}^{3}}=1+\dfrac{3!}{1!2!}\left( \sqrt{2} \right)+\dfrac{3!}{2!1!}\left( 2 \right)+\dfrac{3!}{3!0!}\left( 2\sqrt{2} \right)$.
\[\Rightarrow {{a}^{3}}=1+\dfrac{3}{1}\left( \sqrt{2} \right)+\dfrac{3}{1}\left( 2 \right)+\dfrac{1}{1}\left( 2\sqrt{2} \right)\].
$\Rightarrow {{a}^{3}}=1+3\sqrt{2}+6+2\sqrt{2}$..
We have got the terms 1 and 6 as rational which are 2 ---(4).
From equations (2), (3), (4) we have got a total of $\left( 1+4+2 \right)=7$ rational terms.
∴ We have got a total 7 rational terms in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
The correct option for the given problem is (b).
Note:
We get only one rational if we add all the rational numbers we just obtained. Since they were asked about the number of rational terms present in it, we have taken them separately without adding. We can also take $a=1$ and $b=\sqrt{2}+\sqrt[3]{3}$ for solving this problem in which we just need to expand the powers of ‘b’ only to get the result. Similarly, we can expect problems to find the sum of the rational terms present in this expansion.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Biology: Engaging Questions & Answers for Success
Master Class 11 Physics: Engaging Questions & Answers for Success
Master Class 11 Chemistry: Engaging Questions & Answers for Success
Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
What is the full form of POSCO class 10 social science CBSE
Define Potential, Developed, Stock and Reserved resources
The diagonals of a rhombus are 10cm and 24cm Find the class 10 maths CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
What planets have no moons Which one has only one moon class 10 physics CBSE