Binomial theorem practice · Mixed difficulty

Option A ($\frac{2^{10}+1}{2^{20}}$) is correct. The sum is simplified using the binomial theorem.

Option C (120) is correct because by expanding the binomial and simplifying the summation, the result equals 120.

Option C (-^{100}C_{53}) is correct because the given summation simplifies to (x-1)¹⁰⁰, and the coefficient of x⁵³ is -^{100}C_{53}.

Option A (${\;^{n + 1}}{C_{k + 1}}$) is correct because the sum of the geometric progression can be simplified, and the coefficient of x^k is then found using the binomial theorem.

Option B ($\frac{1}{2}{\;^{20}}{C_{10}}$) is correct because the given series can be simplified using properties of binomial coefficients and the binomial theorem.

Option D (1) is correct because the given expression simplifies to (1-x)^(-1), and the coefficient of x^n in its expansion is 1.

Option C (120) is correct because the coefficient is found using the multinomial theorem and solving for p, q, and r.

Option B ($2^{{m+1}}$ as a factor) is correct because by expressing $(\sqrt{{3}}+1)^{{2 m}}$ as $I+F$ and using its conjugate, we can show that $2^{{m+1}}$ is a factor of the integer just greater than $(\sqrt{{3}}+1)^{{2 m}}$.

Option D (5) is correct because substituting p=5 into the general term formula yields the given numerical coefficient of 4860.

Option D (No element) is correct because the domain of $\sqrt{x^2-1}$ and $\cos^{-1}x$ have no common elements that satisfy the equation.

Option B (108) is correct because the number of integral terms is found by determining the multiples of 6 within the range and adding 1 for the initial term.

Option C ($\frac{{2{a_2}}}{{{a_2} + {a_3}}}$) is correct because the given expression can be simplified using binomial coefficient identities to arrive at this result.

Option B ($\frac{1}{n+1}\left({ }^{2 n+1} C_n\right)$) is correct because the given sum can be transformed using combinatorial identities and then evaluated using the Vandermonde's identity.

Option B ($\frac{1}{{24}}n\left( {{n^2} - 1} \right)\left( {3\;n + 2} \right)$) is correct because it is the correct coefficient of ${x^{n - 2}}$ when expanding the given polynomial.

Option D ($\frac{{{{\left( {n + 1} \right)}^n}}}{{n!}}$) is correct because the given expression simplifies to this form using properties of binomial coefficients.

Option D (ny) is correct because differentiating the binomial expansion and substituting x+y=1 simplifies the expression to ny.

Option D (None of these) is correct because the number of distinct terms in the expansion is $\frac{(n+2)^2}{2}$.

Option B (e) is correct because solving the equation derived from the 7th term of the binomial expansion leads to x = e.

Option A (0, if $n$ is odd) is correct because when $n$ is odd, the term independent of $x$ in the expansion is 0.

Option A (3) is correct because the expression is approximately equal to e, which is between 2 and 3, so the least integer greater than it is 3.

Option B ($\frac{{ - 3}}{8}{x^2}$) is correct because by using binomial expansion and neglecting terms of $x^3$ and higher, the expression simplifies to this value.

Option B ($\frac{{1}}{{2}}(3^{n+2}-2^{n+3}+1)$) is correct because partial fraction decomposition and series expansion are used to find the coefficient of $x^n$.

Option C (-1) is correct because substituting x = 1 into the expression gives the sum of the coefficients, which evaluates to -1.

Option A (2) is correct because the number of terms in the expansion is 101, and the unit's digit of $2^{101}$ is 2.

Option C ($\frac{{1}}{{2}} n$) is correct because by using the property ${}^n C_r = {}^n C_{n-r}$ and summing the series, we can find that $t_n = \frac{n}{2} s_n$, thus $\frac{t_n}{s_n} = \frac{n}{2}$.

Option A ($n \cdot 2^{2 n-1}$) is correct because the sum can be simplified using combinatorial identities to $n \cdot 2^{2 n-1}$.

Option B (6) is correct because solving the given equation leads to x being a complex cube root of unity, allowing calculation of p, and q is found using the properties of powers of 2.

Option C (${2^{14}}$) is correct because the sum of the last eight coefficients in the expansion of ${\left( {1 + x} \right)^{15}}$ is half of the total coefficients, which equals ${2^{14}}$.

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 C (35) is correct because by applying the binomial theorem and simplifying, the remainder when $6^{83}+8^{83}$ is divided by 49 is 35.