If $\frac{1^3+2^3+3^3+\ldots \text{ upto } n \text{ terms}}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \text{ upto } n \text{ terms}} = \frac{9}{5}$,then the value of $n$ is

For any odd integer $n \ge 1$,${n^3} - {(n - 1)^3} + \dots + {( - 1)^{n - 1}}{1^3} = $

Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression with $a_1=7$ and common difference $8$. Let $T_1, T_2, T_3, \ldots$ be such that $T_1=3$ and $T_{n+1}-T_n=a_n$ for $n \geq 1$. Then,which of the following is/are $TRUE$?
$(A) T_{20}=1604$
$(B) \sum_{k=1}^{20} T_k=10510$
$(C) T_{30}=3454$
$(D) \sum_{k=1}^{30} T_k=35610$

The sum of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$ up to $n$ terms is:

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

If $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3, 4, 5, \ldots$,then $f(10)=$