Binomial theorem practice · Mixed difficulty

Option A ($2,2 \pm \sqrt{2}$) is correct because by recognizing the reciprocal relationship between $(7-4\sqrt{3})$ and $(7+4\sqrt{3})$, we can simplify the equation and solve for $x^2-4x+3$, eventually finding the values of $x$.

Option A (-1) is correct because substituting x = 1 into the polynomial yields (1 + 1 - 3)²¹⁴³ = (-1)²¹⁴³ = -1.

Option C (415) is correct because by setting $F=(\sqrt{3}-1)^6$, $R+F$ becomes an integer, and $[R]$ is $416-1 = 415$.

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 ($\frac{{3 n !}}{{2 n ! n !}} 2^n \cdot x^{-n}$) is correct because it is the (n+1)th term from the end in the expansion.

Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because both the sum of even and odd binomial coefficients are equal to 2^(n-1), and this equality explains the assertion.

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 C ($\frac{{{{\left( {n + 1} \right)}^n}}}{{n!}}$) is correct because it is the simplified form of the product of the given expression using binomial coefficients.

Option C (3) is correct because for the term to be independent of x, n must be a multiple of 3.

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 A (7) is correct because the unit digit of 183! is 0 and the unit digit of 3¹⁸³ is 7, so the unit digit of their sum is 7.

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

Option A (1) is correct because by setting $F=(\sqrt{2}-1)^{2 n+1}$, $R-F$ becomes an integer, and $Rf$ simplifies to 1.

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

Option B (17) is correct because when n=1, the expression evaluates to 3(5³) + 2⁴ = 375 + 16 = 391 = 17 * 23, which is divisible by 17.

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!.

Option B ($n - r - 1 = 0$) is correct because equating the binomial coefficients and simplifying leads to the relation n = r + 1, which can be rearranged to n - r - 1 = 0.

Option C (1) is correct because the sum of the coefficients vanishes when a=1, making the expression equal to zero.

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 (2, 3, 4) is correct because solving the equation derived from the A.P. condition yields n=2, 3, or 4.

Option B (120) is correct because given the recursive relation ${a_n} = n{a_{n - 1}}$ and ${a_1} = 1$, calculating successive terms leads to ${a_5} = 120$.

Option B ($5^{\mathrm{n}+1}-4^{\mathrm{n}+1}$) is correct because it is the coefficient of x^n obtained after partial fraction decomposition and series expansion.

Option A (-3/10) is correct because it satisfies the condition where the middle term coefficients are equal in both expansions.

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 is correct. The given sum simplifies to n(n+1)/2, where n is the upper limit of the binomial coefficients.

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.