Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $(2n-1)^2$.

The sum of the first $20$ terms of the series $5+11+19+29+41+\ldots$ is $..........$.

If $\sum_{k=1}^{n} a_k = 6n^3$, then $\sum_{k=1}^{6} \left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to . . . . . . .

The sum of the series $\frac{1}{1-3 \cdot 1^2+1^4} + \frac{2}{1-3 \cdot 2^2+2^4} + \frac{3}{1-3 \cdot 3^2+3^4} + \ldots$ up to $10$ terms is

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 $TRUE$?
$(A) T_{20}=1604$
$(B) \sum_{k=1}^{20} T_k=10510$
$(C) T_{30}=3454$
$(D) \sum_{k=1}^{30} T_k=35610$

If $\alpha_r$ and $\beta_r$ (where $\alpha_r < \beta_r$) are the roots of the quadratic equation $x^2 - r^2(r + 1)x + r^5 = 0$,then find the value of $\sum_{r=1}^{n} (3\alpha_r + 2\beta_r)$.