Find the $9^{\text{th}}$ term in the sequence whose $n^{\text{th}}$ term is given by $a_{n} = (-1)^{n-1} n^{3}$.

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

The sum $1+3+11+25+45+71+\ldots$ up to $20$ terms is equal to:

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

The expression for $a_n$ which satisfies $a_0=0, a_1=1$ and $a_n=a_{n-1}+a_{n-2}, \forall n \in N -\{0,1\}$ is:

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$