For $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$ to exceed $1.01$,the maximum value of $n$ is

The sum of the infinite series $\frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + \dots \infty$ is equal to :-

Let $a_n$ denote the number of all $n$-digit positive integers formed by the digits $0, 1$ or both such that no consecutive digits in them are $0$. Let $b_n$ be the number of such $n$-digit integers ending with digit $1$ and $c_n$ be the number of such $n$-digit integers ending with digit $0$.
$1.$ Which of the following is correct?
$(A)$ $a_{17} = a_{16} + a_{15}$
$(B)$ $c_{17} \neq c_{16} + c_{15}$
$(C)$ $b_{17} \neq b_{16} + c_{16}$
$(D)$ $a_{17} = c_{17} + b_{16}$
$2.$ The value of $b_6$ is
$(A)$ $7$ $(B)$ $8$ $(C)$ $9$ $(D)$ $11$
Give the answer for question $1$ and $2$.

Let $u_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$,for $n = 0, 1, 2, ...$. Then which of the following is true?

Find the sum to $n$ terms of the series $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$

The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots$ is equal to