If $x$ is positive real number and the first negative term in the expansion of $(1+x)^{\frac{27}{5}}$ is $t_k$, then $k=$

The coefficient of $x^{10}$ in the expansion of $\left(x+\frac{2}{x}-5\right)^{12}$ is

Let $S_1=\sum\limits_{j=1}^{10} j(j-1) \cdot{ }^{10} C_j, S_2=\sum\limits_{j=1}^{10} j \cdot{ }^{10} C_j$ and

$$ S_3=\sum\limits_{j=1}^{10} j^2 \cdot{ }^{10} C_j $$

Assertion (A) $S_3=55 \times 2^9$

Reason (R) $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$

If $y=\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots+\infty$, then