Mock Test 2 — Binomial Theorem

If the coefficient of $x^2$ in the expansion of $(1 + cx + x^2)(1 - x)^{10}$ is exactly zero, then the value of the scalar constant $c$ is:

The value of the double sum expression $\sum_{0 \le i < j \le n} \sum \binom{n}{i}\binom{n}{j}$ is equal to:

The coefficient of $x^5$ in the expanded infinite series of $\frac{1}{(1-x)^3}$ for $|x| < 1$ is:

If $S = \sum_{r=1}^{n} r^2 \binom{n}{r}$, then the series sum can be simplified using derivative transformations to:

The number of distinct non-negative integer solutions to the multinomial constraint equation $n_1 + n_2 + n_3 = 6$ is exactly:

The value of the sum series $\binom{30}{0}\binom{30}{10} + \binom{30}{1}\binom{30}{11} + \dots + \binom{30}{20}\binom{30}{30}$ is equal to:

The value of the continuous product ratio $\prod_{r=1}^{n} \frac{C_r}{C_{r-1}}$ can be simplified to:

If the expression $x$ is so small that its square and higher powers can be ignored, the value of $\frac{(1+x)^{3/2} - (1-x/2)^3}{(1-x)^{1/2}}$ simplifies to:

The single Numerically Greatest Term (NGT) in the expansion of $(1 + 3x)^{10}$ when $x = \frac{1}{3}$ occurs at which term position?

The remainder when the big number $3^{2026}$ is divided by $8$ is:

Find the value of the subfactorial derangement constant $D_4$, which tracks the ways to completely mix up 4 letters.

Find the coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^4$.

If $\sum_{r=1}^{5} r \cdot \binom{5}{r} = A$, find the numerical value of $A$.

Assertion (A): The maximum value among the binomial coefficients $\binom{n}{r}$ always occurs at the central coordinate index $\binom{n}{\lfloor n/2 \rfloor}$.
Reason (R): Binomial coefficients grow symmetrically from both ends of Pascal's triangle, reaching a single peak or a pair of equal peak values in the middle.

Assertion (A): The double sum expression $\sum_{i=0}^n \sum_{j=0}^n C_i C_j$ expands cleanly to the square value $(2^n)^2 = 2^{2n}$.
Reason (R): Since the two summation indices $i$ and $j$ operate completely independently of each other, the double sum can be factored into the product of two separate single sums.