$\sum_{k=1}^{2n+1} (-1)^{k-1} \cdot k^2$ is equal to

Let $\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots$ up to $10$ terms and $\beta = \sum_{n=1}^{10} n^4$. If $4\alpha - \beta = 55k + 40$,then $k$ 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 $S_n = \sum_{k=1}^n (-1)^{k-1} \cdot k^2$ for $n \geq 1$. Given that $S_{2n} = -n(2n+1)$ for $n = 1, 2, 3, \ldots$,then $S_{77} =$

The numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

Find the sum to $n$ terms of the series: $5+11+19+29+41 + \ldots$