$\sum\limits_{r = 0}^{100} {({r^2} + 4r + 4)(r + 1)!}$ is equal to :-

If ${x_1}, {x_2}, {x_3}$ as well as ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

The value of $a^{\log_b x}$,where $a = 0.2$,$b = \sqrt{5}$,and $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ to $\infty$ is:

There are two such pairs of non-zero real values of $a$ and $b$,i.e.,$(a_1, b_1)$ and $(a_2, b_2)$,for which $2a+b, a-b, a+3b$ are three consecutive terms of a $G.P.$. Then the value of $2(a_1b_2 + a_2b_1) + 9a_1a_2$ is-

The value of $(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots$ up to $51^{\text{th}}$ term) + $(1! - 2! + 3! - \dots$ up to $51^{\text{th}}$ term) is equal to

If the $n^{th}$ term of the geometric progression $5, - \frac{5}{2}, \frac{5}{4}, - \frac{5}{8}, \dots$ is $\frac{5}{1024}$,then the value of $n$ is: