Binomial theorem practice · Mixed difficulty

Option D (an irrational number) is correct because the expression simplifies to $2\sqrt{3}$ multiplied by an integer, making it an irrational number.

Option A ($a_p=a_q$) is correct because the binomial coefficients ${}^{p+q}C_p$ and ${}^{p+q}C_q$ are equal due to the property ${}^nC_r = {}^nC_{n-r}$.

Option A (0) is the correct answer because the sum evaluates to zero when n is an odd natural number.

Option D ($400\,{\,^{38}}{C_{20}}$) is correct because the given summation can be simplified using combinatorial identities to a form involving ${ }^{38} C_{20}$.

Option B is correct because the coefficient of x¹⁰ in the expansion of (3x² - 1/x²)¹⁵ is - (15! / (10! 5!)) * 3¹⁰, which is found using the binomial theorem.

Option D ($\frac{{20!}}{{5!15!}} - \frac{{15!}}{{5!10!}}$) is correct because it represents the coefficient of x¹⁵ in the expansion of (1+x)²⁰ minus the term where both binomials contribute to x¹⁵.

Option C (64) is correct because $9^7 + 7^9$ is divisible by 64.

Option B (${ }^9 C_6\left(\frac{{1}}{{6}}\right)^3$) is correct because it represents the constant term in the expansion, obtained by setting the power of x to zero and simplifying.

Option A ($(n+1) 2^{n+1}$) is the correct answer because the summation can be simplified by separating the terms and using the binomial theorem.

Option B ($r+1$) is correct because the given expression is the expansion of $(1-x)^{-2}$, and the coefficient of $x^r$ is $r+1$.

Option C (${ }^{n+1} C_{m+1}$) is correct because the sum of binomial coefficients can be simplified using the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$ repeatedly.

Option D (${\;^n}{C_4} + {\;^n}{C_2} + {\;^n}{C_1} \times {\;^n}{C_2}$) is correct because it represents the sum of the products of binomial coefficients that result in the $x^4$ term when expanding the given expression.

Option A (${ }^{16} C_9$) is correct because the given expression simplifies to a single binomial expansion, and the coefficient of x⁶ is ${ }^{16} C_7$ which equals ${ }^{16} C_9$.

Option B ($(2 {a}+{nb}) 2^{{n}-1}$) is correct because the given series can be simplified using binomial coefficient properties and summation techniques.

Option A (0) is correct because the given expression represents the coefficient of x²⁵ in (1-x)⁴⁰(1+x)⁴⁰ which simplifies to (1-x²)⁴⁰, and thus the coefficient of x²⁵ is 0.

Option B (924) is correct because when the sum of coefficients in ${\left( {x + y} \right)^n}$ is 4096, n is 12, and the greatest coefficient is ${\;^{12}}{C_6}$ which equals 924.

Option B (-144) is correct because it is the result of calculating the coefficient of $x^7$ after factoring and expanding the given expression.

Option A (${\;^{60}}{C_{40}}\;\;$) is correct because ${(1 + 3x + 3{x^2} + {x^3})^{20}}$ simplifies to ${(1 + x)^{60}}$, and the coefficient of ${x^{20}}$ is ${\;^{60}}{C_{20}}$ which is equal to ${\;^{60}}{C_{40}}$.

Option C ($1-\frac{5}{8} x$) is correct because it is the simplified form of the given expression when higher powers of x are neglected.

Option A ($2^{2n}$) is correct because the summation can be simplified to the square of the sum of binomial coefficients, which equals $2^{2n}$.

Option D $\left(\begin{array}{c}16 \\ 7\end{array}\right)$ is correct because it represents the number of ways to distribute the marks with the condition that each equation receives at least 2 marks.

Option C (141) is correct because the constant term in the expansion is obtained by summing the products of coefficients that result in $x^0$.

Option A (4) is correct because after simplification, the coefficient of $x^{13}$ in the expansion is found to be 4.

Option D (430) is correct because it represents the number of ways to distribute the candies, calculated using the multinomial theorem and considering the given constraints on the number of candies each child can receive.

Option A ($a b+1=0$) is correct because equating the coefficients of $x^8$ and $x^{-8}$ leads to the relation $ab = -1$.

Option D (2) is correct because the greatest value of the term independent of x is achieved when sin(2α)=1, leading to a=2 after solving the equation.

Option B (23) is correct because, after comparing the given series with the binomial expansion and solving for α, we find that α^2 + 4α equals 23.

Option B ($\frac{{15}}{{8}}$) is correct because it is the result of calculating the coefficient of $x^3$ in the series expansion of the given function.

Option B (100) is correct because by comparing the given product with the binomial expansion, we can deduce that n = 100.

Option C ($\frac{1}{2}(3^{50}+1)$) is correct because it uses the property of binomial expansions to find the sum of coefficients of integral powers of x.