The minimum value of $n$ for which $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2} < 1.01$ is

If the set of natural numbers is partitioned into subsets $S_1 = \{1\}, S_2 = \{2, 3\}, S_3 = \{4, 5, 6\}$ and so on,then the sum of the terms in $S_{50}$ is

$A$ person wants to climb a $n$-step staircase using one step or two steps at a time. Let $C_n$ denote the number of ways of climbing the $n$-step staircase. Then $C_{18} + C_{19}$ equals

The sum of the series $\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\ldots + \frac{30^{3}-29^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$ is equal to:

For a series $S = 1 - 2 + 3 - 4 + \dots$ up to $n$ terms,
Statement-$1$: The sum of the series is always dependent on the value of $n$,i.e.,whether it is even or odd.
Statement-$2$: The sum of the series is $-\frac{n}{2}$ when the value of $n$ is any even integer.

For all $n \in N$,if $1^3+2^3+3^3+\ldots+n^3 > x$,then a value of $x$ among the following is