Binomial theorem practice · Medium

Option D (None of these) is correct because the sum of the last eight terms is calculated using binomial coefficients and the properties of binomial expansions, which results in a different expression than the other options.

Option D ($\left( {2r + 1} \right){2^r}$) is correct because the coefficient of ${x^r}$ in the expansion of $\frac{{1 + 2x}}{{{{\left( {1 - 2x} \right)}^2}}}$ is $\left( {2r + 1} \right){2^r}$.

Option A (97) is correct because by finding the number of rational terms in the expansion and subtracting it from the total number of terms, we get the number of irrational terms.

Option B ($2^{{n}-2}\left(n^2+5 n+4\right)$) is correct because it is the result of applying differentiation and multiplication by x to the binomial expansion and then substituting x=1.

Option C (0) is correct because the summation can be split into two parts, both of which evaluate to zero using binomial identities.

Option C (0) is correct because when $r$ is not a multiple of $p+1$, the coefficient of $x^r$ in the expansion of $(1-x^{p+1})^n$ is 0.

Option B (5456) is correct because it represents the number of non-negative integral solutions to the inequality, calculated using combinations with repetition.

Option B $\left( {1, - \frac{{35}}{{24}}} \right)$ is correct because it is obtained by using binomial approximation and comparing coefficients.

Option B ($\frac{{2^{n-1}}}{{n!}}$) is correct because the given expression can be rewritten in terms of binomial coefficients, and then simplified to the sum of odd-indexed coefficients divided by n!.

The correct answer is C (202) because simplifying the expression leads to (1+x)(1-x³)¹⁰⁰, which has 202 terms after expansion.

Option D is correct because the coefficient is found using the multinomial theorem, and the correct coefficient is $\frac{{9 !}}{{3 ! 4 !}}$.

Option C (4) is correct because when $5^{2222}$ is divided by 7, the remainder is 4, which can be found using modular arithmetic and properties of exponents.

Option B (9) is correct because f(n) can be rewritten in a form that is always divisible by 9 for all natural numbers n.

Option A ($2^n$) is correct because the given expression is equivalent to the binomial expansion of (1+1)^n, which equals 2^n.

Option B (1) is correct because substituting x=1 into the polynomial and equating the coefficients leads to the solution α = 1.

Option B ($\frac{n(n+1)}{2}$) is correct because the given expression simplifies to the sum of the first n natural numbers, which is $\frac{n(n+1)}{2}$.

Option A ($\frac{{{{2^{mn}} - 1}}}{{{{2^{mn}}\left( {{{2^n} - 1}} \right)}}}$) is the correct answer because it represents the sum of the given series after simplification using binomial theorem and geometric progression.

Option C (2) is correct because the expression can be approximated using the binomial theorem, and the result is slightly greater than 1 but less than 2.

Option C ($2^{12}$) is correct because the sum of numerical coefficients is found by substituting x=1 and y=1 into the expression.

Option D (32) is correct because the third term in the binomial expansion is given by ${}^nC_2 (x^2)^{n-2} (1/4)^2$, and solving for n yields 32.

Option B (${4^n}$) is correct because by substituting $x = i$ and $x = -i$ into the given expression and using complex number properties, we can simplify the expression to ${4^n}$.

Option C ($(50,101)$) is correct because it is obtained by calculating the sum of coefficients of odd powers, the middle term, and then simplifying the given expression.

Option D ($\frac{1}{{1 - f}} - f$) is correct because it is derived from the properties of conjugate surds and integer/fractional parts.

Option A (1) is correct because by considering the conjugate and using the properties of greatest integer function, we can simplify the expression to 1.

Option A (3) is correct because substituting x = -1 into the given expression gives (-3)²¹⁴⁵, and the last digit of this number is 3.

Option C ($2^{n-1} \sin n x$) is correct because the summation simplifies to this expression using trigonometric identities and the binomial theorem.

Option D is correct because setting the sum of the 5th and 6th terms to zero and simplifying the binomial coefficients leads to the given ratio.

Option C (-5) is correct because it is the constant term obtained after expanding the given expression and simplifying.

Option D ($a_{{n-3}}=a_{{n+3}}$) is correct because binomial coefficients are symmetric, meaning ${ }^{2n}C_{{n-3}} = { }^{2n}C_{{n+3}}$.

Option B ($n(n+3) \cdot 2^{n-3}$) is correct because it is derived by simplifying the double summation involving binomial coefficients.