Explanation
$${\textbf{Step - 1: Determine the Middle term}}$$
$${\text{Middle term = }}\dfrac{{\text{N}}}{2} + 1 = \dfrac{{2{\text{n}}}}{2} + 1$$
$$ = \left( {{\text{n + 1}}} \right){\text{th term}}$$
$${\text{The middle term of }}{\left( {{\text{1 + x}}} \right)^{40}}{\text{ is }}\left( {\dfrac{{40}}{2} + 1} \right){\text{ i}}{\text{.e 2}}{{\text{1}}^{th}}{\text{ term in that expansion}}$$
$${{\text{T}}_{21}} = {{\text{ }}^{40}}{{\text{C}}_{20}}{{\text{x}}^{40 - 20}} = {{\text{ }}^{40}}{{\text{C}}_{20}}{{\text{x}}^{20}}$$
$${\text{Therefore the coefficient of the middle term is}}{{\text{ }}^{40}}{{\text{C}}_{20}}$$
$$\therefore {{\text{ }}^{40}}{{\text{C}}_{20}} = \dfrac{{40!}}{{20!\left( {40 - 20} \right)!}} = \dfrac{{40!}}{{20!20!}}$$
$${\textbf{Step - 2 : Simplify the term}}{\text{.}}$$
$${\text{Dividing by 20!;and separating the multiples of two in the powers of two we get,}}$$
$$\therefore {\text{Middle term = }}\dfrac{{\left( {1.3.5.7.....39} \right){2^{20}}}}{{20!}}$$
$${\textbf{Hence, Correct answer will be }}\dfrac{{\left( {1.3.5.7.....39} \right){2^{20}}}}{{20!}}$$
(i) $$(5^{\frac{1}{6}}+7^{\frac{1}{9}})^{1824}$$
$$T_{r+1}=^{1824}C_r (5)^{\frac{1824-r}{6}} 7^{\frac{r}{9}}$$
For integer terms, $$r$$ should be multiple of $$9$$.
For $$r=18,36,54,72,.....1818$$, terms comes as integer.
This is an A.P.
$$1818=18+(n-1)18$$
$$\Rightarrow n=101$$
Also, for $$r=0$$ , we would get an integer
So, total number of terms which gives integer values are $$101+1=102.$$
So, $$\lambda=102$$
So, $$\lambda$$ is divisible by $$2,3,17$$
(ii) $$(5^{\frac{1}{6}}+2^{\frac{1}{8}})^{1824}$$
$$T_{r+1}=^{100}C_r (5)^{\frac{100-r}{6}}2^{\frac{r}{8}}$$
For rational terms, $$r$$ should be multiple of $$8$$.
For $$r=16,40,64,88$$, terms comes as rational.
So, number of rational terms are $$4$$.
So, $$\lambda=4$$
which is divisble by $$2$$.
(iii) $$(3^{\frac{1}{4}}+4^{\frac{1}{3}})^{99}$$
$$T_{r+1}=^{99}C_r (3)^{\frac{99-r}{64}}4^{\frac{r}{3}}$$
For rational terms, $$r$$ should be multiple of $$3$$.
For $$r=3,15,27,.....97$$, terms comes as rational.
This is an AP
$$97=3+(n-1)12$$
$$\Rightarrow n=8$$
For $$r=99$$ also, there is a rational value
So, number of rational terms are $$8+1=9$$
Now, number of irrational terms = total number of terms -rational number of terms
$$=99+1-9=91$$
So, $$\lambda=91$$
which is divisble by $$7,13$$.
Hence, option A is the correct answer.
$$(1+x)^{2n+2}$$. The middle term will be $$(\dfrac {N}{2}+1)$$ th term. $$=\dfrac {2n+2}{2}+1$$ th term $$=(n+2)^{th}$$ term Hence coefficient of the middle term will be $$\:^{2n+2}C_{n+1}$$ $$=p$$ For $$(1+x)^{2n+1}$$. The middle terms will be $$(\dfrac {N+1}{2}+1)$$ and $$(\dfrac {N+1}{2})$$ terms $$=\dfrac {2n+2}{2}+1$$ th term and $$\dfrac {2n+2}{2}^{th}$$ term. $$=(n+2)^{th}$$ term and $$(n+1)^{th}$$ term. Hence coefficient of the middle terms will be $$\:^{2n+1}C_{n+1}=q$$ and $$\:^{2n+1}C_{n}=r$$ By the properties of binomial coefficients. $$\:^{2n+1}C_{n+1}+\:^{2n+1}C_{n}$$ $$=\:^{2n+2}C_{n+1}$$ Hence $$q+r=p$$
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