Question

Number of rational terms in the expansion of ${{\left( \sqrt{2}+\sqrt{3}+\sqrt[3]{5} \right)}^{20}}$is a two digit number $ab$ then $a+b=$ $A.3$
B.6$C.8$
D.9$$$$

Answer 1

We expand the given expression ${{\left( \sqrt{2}+\sqrt{3}+\sqrt[3]{5} \right)}^{20}}$ by trinomial expansion and we find the general term is ${}^{20}{{C}_{k,l,m}}{{\left( \sqrt{2} \right)}^{k}}{{\left( \sqrt{3} \right)}^{l}}{{\left( \sqrt[3]{5} \right)}^{m}}$ where $k,l,m$ are non-negative integers . We also find that the term is rational when $k,l$ are even numbers and $m$ is a multiple of 3. We find all such ordered triples $\left( k,l,m \right)$ such that $k+l+m=20$ and the number of ordered is the number of rational terms. $$$$

Complete step-by-step solution
We know from trinomial expansion of three real numbers $a,b,c$ with non-negative integers $k,l,m$ that

k,l,m \\\ \end{smallmatrix}}{\left( {{T}_{n}} \right){{a}^{k}}{{b}^{l}}{{c}^{m}}}$$ Here ${{T}_{n}}$ is the binomial coefficient of the ${{n}^{\text{th}}}$ in the expansion which is given by $${{T}_{n}}={}^{n}{{C}_{k,l,m}}=\dfrac{n!}{k!l!m!}$$ We are given the expression in the question as , $${{\left( \sqrt{2}+\sqrt{3}+\sqrt[3]{5} \right)}^{20}}$$ Let us expand the above expression by trinomial expansion as there are three terms $a=\sqrt{2},b=\sqrt{3},c=\sqrt[3]{5}$. Here we have $n=20.$ So we have for some non-negative integers $k,l,m$ , $$\begin{aligned} & {{\left( \sqrt{2}+\sqrt{3}+\sqrt[3]{5} \right)}^{20}}=\sum\limits_{k+l+m=20}{{}^{20}{{C}_{k,l,m}}{{\left( \sqrt{2} \right)}^{k}}{{\left( \sqrt{3} \right)}^{l}}{{\left( \sqrt[3]{5} \right)}^{m}}} \\\ & \Rightarrow {{\left( \sqrt{2}+\sqrt{3}+\sqrt[3]{5} \right)}^{20}}=\sum\limits_{k+l+m=20}{{}^{20}{{C}_{k,l,m}}{{\left( 2 \right)}^{\dfrac{k}{2}}}{{\left( 3 \right)}^{\dfrac{l}{2}}}{{\left( 5 \right)}^{\dfrac{m}{3}}}}...\left( 1 \right) \\\ \end{aligned}$$ We see that the exponents $k,l,m$ have to satisfy the condition $k+l+m=20$. We observe the expansion and the exponents occurring in them as $\dfrac{k}{2},\dfrac{l}{2},\dfrac{m}{3}$. If the term has to be rational then $k,l$ have to be multiple of 2 in other words an even number and $m$ has to be multiple of $3$. So the possible choices for $k$ are $$k=0,2,4,6,8,10,12,14,16,18,20$$ The possible choices of $l$ are $$l=0,2,4,6,8,10,12,14,16,18,20$$ The possible choices of $m$ are $$m=0,3,6,9,12,15,18$$ So we have to choose $k,l$ in a way such that $m=20-\left( k+l \right)$ is exactly divisible by 3. So $k+l$ is an even number then $m=20-\left( k+l \right)$ is an even number and hence even multiple of 3. So choices of $m$ are $$m=0,6,12,18$$ So such the ordered triples $\left( k,l,m \right)$ which satisfies $k+l+m=20$ are$$\begin{aligned} & k=0,2,4,6,8,10,12,14,16,18,20 \\\ & l=0,2,4,6,8,10,12,14,16,18,20 \\\ & m=0,6,12,18 \\\ & \left( k,l,m \right)= \left( 0,2,18 \right),\left( 0,8,12 \right),\left( 0,14,6 \right),\left( 0,20,0 \right),\left( 2,0,18 \right),\left( 2,6,12 \right),\left( 2,12,6 \right), \\\ & \left( 2,18,0 \right),\left( 4,4,12 \right),\left( 4,10,6 \right),\left( 4,16,0 \right),\left( 6,2,12 \right),\left( 6,8,6 \right),\left( 6,14,0 \right),\left( 8,0,12 \right), \\\ & \left( 8,6,6 \right),\left( 8,12,0 \right),\left( 10,4,6 \right),\left( 12,2,6 \right),\left( 12,8,0 \right),\left( 14,0,6 \right),\left( 14,6,0 \right),\left( 16,4,0 \right), \\\ & \left( 18,2,0 \right),\left( 20,0,0 \right),\left( 0,20,0 \right) \\\ \end{aligned}$$ So the number of rational terms is 26 which is given in the question as $ab.$So the digit at unit place is $b=6$ and the digit at tenth place is $a=2$. So we have $$a+b=2+6=8$$ **So the correct option is C.** **Note:** We can alternatively find the number of rational terms by taking choices for $l,m$ such that $l$ is an even number and $m$ is multiple of 3 and $l+m\le 20$. The general term in the binomial expansion is $^{n}{{C}_{k}}{{a}^{k}}{{b}^{n-k}}$ and the general term in the multinomial expansion with $m$ terms is ${}^{n}{{C}_{{{k}_{1}},{{k}_{2}},..,{{k}_{m}}}}\prod\limits_{i=1}^{m}{{{x}_{i}}^{{{k}_{i}}}}$ where the nonnegative integers ${{k}_{1}}+{{k}_{2}}+...+{{k}_{n}}=20$.