$$\begin{aligned}\sum_{0\leq k\lt n}(a_{k+1}-a_k)b_k&=\sum_{0\leq k\lt n}\big(a_{k+1}b_{k+1}-a_kb_k-a_{k+1}(b_{k+1}-b_k)\big)\\&=\sum_{0\leq k\lt n}(a_{k+1}b_{k+1}-a_kb_k)-\sum_{0\leq k\lt n}a_{k+1}(b_{k+1}-b_k)\\&=a_nb_n-a_0b_0-\sum_{0\leq k\lt n}a_{k+1}(b_{k+1}-b_k),\quad\text{for $n\geq0$.}\end{aligned}$$
If $x=p(k)$ then $x+c=k+\big((-1)^k+1\big)c$. Observe that $\big((-1)^k+1\big)c$ is even, $x+c$ must have the same parity as $k$. This gives $(-1)^k=(-1)^{x+c}$ and $k=x-(-1)^{x+c}c$. Conversely, this value of $k$ yields $x=p(k)$.
The sum is a special case of the general recurrence
$$\begin{aligned}R_0&=\alpha;\\R_n&=R_{n-1}+(-1)^n(\beta+\gamma n+\delta n^2),\quad\text{for $n>0$,}\end{aligned}$$
whose solution would be
$$R(n)=A(n)\alpha+B(n)\beta+C(n)\gamma+D(n)\delta.$$
Setting $R_n=1$ yields $A(n)=1$.
Setting $R_n=(-1)^n$ yields $A(n)+2B(n)=(-1)^n$.
Setting $R_n=(-1)^nn$ yields $-B(n)+2C(n)=(-1)^nn$.
Setting $R_n=(-1)^nn^2$ yields $-C(n)+2D(n)=(-1)^nn^2$.
Therefore, $D(n)=(-1)^n(n^2+n)/2$.
For this sum, we have $\delta=1$; hence
$$\sum_{k=0}^n(-1)^kk^2=(-1)^n(n^2+n)/2.$$
$$\begin{aligned}\sum_{k=1}^nk2^k&=\sum_{1\leq j\leq k\leq n}2^k\\&=\sum_{1\leq j\leq n}\sum_{j\leq k\leq n}2^k\\&=\sum_{1\leq j\leq n}(2^{n+1}-2^j)\\&=n2^{n+1}-(2^{n+1}-2)\\&=(n-1)2^{n+1}-2.\end{aligned}$$
$$\begin{aligned}\sum_{k=1}^nk^3+\sum_{k=1}^nk^2&=2\sum_{1\leq j\leq k\leq n}jk\\\sum_{k=1}^nk^3+\sum_{k=1}^nk^2&=\left(\sum_{1\leq k\leq n}k\right)^2+\sum_{1\leq k\leq n}k^2\\\sum_{k=1}^nk^3&=\left(\sum_{1\leq k\leq n}k\right)^2\\\sum_{k=1}^nk^3&=n^2(n+1)^2/4.\end{aligned}$$
If no denominator is zero,
$$\begin{aligned}x^{\underline m}/(x-n)^{\underline m}&=x^{\underline n}/(x-m)^{\underline n}\\x^{\underline m}(x-m)^{\underline n}&=x^{\underline n}(x-n)^{\underline m},\end{aligned}$$
the latter is applicable to the law of exponents.
$$\begin{aligned}x^{\overline m}&=x(x+1)\cdots(x+m-1)\\&=(-1)^m(-x)(-x-1)\cdots(-x-m+1)\\&=(-1)^m(-x)^{\underline m}.\end{aligned}$$
$$\begin{aligned}x^{\overline m}&=x(x+1)\cdots(x+m-1)\\&=(x+m-1)\cdots(x+1)x\\&=(x+m-1)^{\underline m}.\end{aligned}$$
$$\begin{aligned}x^{\overline m}&=x(x+1)\cdots(x+m-1)\\&=1/(x-1)^{\underline{-m}}.\end{aligned}$$
The second line is similar.
If $\sum_{k\in K}a_k$ is absolutely convergent, so are $\sum_{k\in K}(\Re a_k)^+$, $\sum_{k\in K}(\Re a_k)^-$, $\sum_{k\in K}(\Im a_k)^+$ and $\sum_{k\in K}(\Im a_k)^-$.
For all finite subset $F\subsetneq K$, there must be a bounding constant $B$ such that
$$\sum_{k\in F}|a_k|\leq\sum_{k\in F}\big((\Re a_k)^++(\Re a_k)^-+(\Im a_k)^++(\Im a_k)^-\big)\leq B.$$
Conversely, with the fact that $(\Re z)^+,(\Re z)^-,(\Im z)^+,(\Im z)^-\leq|z|$, we know that $\sum_{k\in K}(\Re a_k)^+$, $\sum_{k\in K}(\Re a_k)^-$, $\sum_{k\in K}(\Im a_k)^+$ and $\sum_{k\in K}(\Im a_k)^-$ are each absolutely convergent, so is $\sum_{k\in K}a_k$.
Multiply both sides by $2^{n-1}/n!$, we get
$$\begin{aligned}\frac{2^n}{n!}T_n&=\frac{2^{n-1}}{(n-1)!}T_{n-1}+3\cdot2^{n-1}\\T_n&=\frac{n!}{2^n}\left(T_0+\sum_{k=1}^n3\cdot2^{n-1}\right)\\T_n&=3\cdot n!+\frac{n!}{2^{n-1}}.\end{aligned}$$
$$\begin{aligned}\sum_{k=0}^nkH_k&=\sum_{k=0}^n\big((k+1)H_{k+1}-(n+1)H_{n+1}\big)\\\sum_{k=0}^nkH_k&=\sum_{k=0}^nkH_{k+1}+\sum_{k=0}^nH_{k+1}-(n+1)H_{n+1}\\(n+1)H_{n+1}&=\sum_{k=0}^n\frac k{k+1}+\sum_{k=0}^nH_{k+1}\\(n+1)H_{n+1}&=\sum_{k=0}^n(H_k+1)\\(n+1)(H_n-1)&=\sum_{k=0}^nH_k.\end{aligned}$$
$$\begin{aligned}\sum_{k=0}^n(-1)^{n-k}&=\sum_{k=0}^n(-1)^{n-k-1}+(-1)^n+1\\2\sum_{k=0}^n(-1)^{n-k}&=(-1)^n+1\\S_n&=\begin{cases}1,&\text{$n$ is even;}\\0,&\text{$n$ is odd.}\end{cases}\end{aligned}$$
$$\begin{aligned}\sum_{k=0}^n(-1)^{n-k}k&=\sum_{k=0}^n-(-1)^{n-k}(k+1)+n+1\\2\sum_{k=0}^n(-1)^{n-k}k&=-\sum_{k=0}^n(-1)^{n-k}+n+1\\2T_n&=-S_n+n+1\\T_n&=\begin{cases}n/2,&\text{$n$ is even;}\\(n+1)/2,&\text{$n$ is odd.}\end{cases}\end{aligned}$$
If we did the first step for $T_n$ differently, we could find out that $\sum_{k=0}^n(-1)^{n-k}(2k+1)=n+1$, which would be useful later.
$$\begin{aligned}\sum_{k=0}^n(-1)^{n-k}k^2&=\sum_{k=0}^n-(-1)^{n-k}(k^2+2k+1)+n^2+2n+1\\2\sum_{k=0}^n(-1)^{n-k}k^2&=-\sum_{k=0}^n(-1)^{n-k}(2k+1)+n^2+2n+1\\2\sum_{k=0}^n(-1)^{n-k}k^2&=n^2+n\\U_n&=n(n+1)/2.\end{aligned}$$
$$\begin{aligned}&\sum_{1\leq j\lt k\leq n}(a_jb_k-a_kb_j)(A_jB_k-A_kB_j)\\&\quad=\sum_{1\leq j\lt k\leq n}(a_jb_kA_jB_k+a_kb_jA_kB_j)-\sum_{1\leq j\lt k\leq n}(a_jb_kA_kB_j+a_kb_jA_jB_k)\\&\quad=\sum_{1\leq j,k\leq n}a_jb_kA_jB_k-\sum_{1\leq k\leq n}a_kb_kA_kB_k-\sum_{1\leq j,k\leq n}a_jb_kA_kB_j+\sum_{1\leq k\leq n}a_kb_kA_kB_k\\&\quad=\sum_{1\leq j,k\leq n}a_jb_kA_jB_k-\sum_{1\leq j,k\leq n}a_jb_kA_kB_j\\&\quad=\left(\sum_{k=1}^na_kA_k\right)\left(\sum_{k=1}^nb_kB_k\right)-\left(\sum_{k=1}^na_kB_k\right)\left(\sum_{k=1}^nb_kA_k\right).\end{aligned}$$
$$\begin{aligned}\sum_{k=1}^n\frac{2k+1}{k(k+1)}&=\sum_{k=1}^n(2k+1)\left(\frac1{k}-\frac1{k+1}\right)\\&=\sum_{k=1}^n\frac1k+\sum_{k=1}^n\frac1{k+1}\\&=2H_n-\frac n{n+1}.\end{aligned}$$
Let $u=2k+1$, then $\Delta v=(k-1)^{\underline{-2}}$, $\Delta u=2$ and $v=-(k-1)^{\underline{-1}}$.
$$\begin{aligned}\sum_{k=1}^n\frac{2k+1}{k(k+1)}&=-\frac{2k+1}k\Bigg|_1^{n+1}+\sum _{1\leq k\lt n+1}2k^{\underline{-1}}\\&=2H_n-\frac n{n+1}.\end{aligned}$$
Sum by parts to evaluate the general form $\sum_{0\leq k\lt n}H_kk^{\underline m}$, let $u=H_k$, then $\Delta v=k^{\underline m}$, $\Delta u=k^{\underline{-1}}$ and $v=k^{\underline{m+1}}/(m+1)$.
$$\begin{aligned}\sum_{0\leq k\lt n}H_kk^{\underline m}&=H_k\frac{k^{\underline{m+1}}}{m+1}\Bigg|_0^n-\sum _{0\leq k\lt n}\frac{k^{\underline m}}{(m+1)^2}\\&=\frac{0^{\underline{m+1}}}{(m+1)^2}+\left(\frac{H_n}{m+1}-\frac1{(m+1)^2}\right).\end{aligned}$$
In this case, we have $m=-2$, so the sum is $1-(H_n+1)/(n+1)$.
$$\begin{aligned}&\prod_{k\in K}a_k^c=\left(\prod_{k\in K}a_k\right)^c;&&\text{(distributive law)}\\&\prod_{k\in K}a_kb_k=\left(\prod_{k\in K}a_k\right)\left(\prod_{k\in K}b_k\right);&&\text{(associative law)}\\&\prod_{k\in K}a_k=\prod_{p(k)\in K}a_{p(k)};&&\text{(commutative law)}\\&\prod_{k\in K}a_k=\prod_ka_k^{[k\in K]};&&\text{(Iverson’s convention)}\\&\prod_{\substack{j\in J\\k\in K}}a_{j,k}=\prod_{j\in J}\prod_{k\in K}a_{j,k};&&\text{(interchanging the order)}\\&\prod_{\substack{j\in J\\k\in K}}a_j^{b_k}=\left(\prod_{j\in J}a_j\right)^{\sum_{k\in K}b_k}.&&\text{(general distributive law)}\end{aligned}$$
$$\begin{aligned}\left(\prod_{1\leq j\leq k\leq n}a_ja_k\right)^2&=\left(\prod_{1\leq j,k\leq n}a_ja_k\right)\left(\prod_{1\leq j=k\leq n}a_ja_k\right)\\\left(\prod_{1\leq j\leq k\leq n}a_ja_k\right)^2&=\left(\prod_{1\leq k\leq n}a_k\right)^n\left(\prod_{1\leq k\leq n}a_k\right)^n\left(\prod_{1\leq k\leq n}a_k^2\right)\\\prod_{1\leq j\leq k\leq n}a_ja_k&=\left(\prod_{1\leq k\leq n}a_k\right)^{n+1}.\end{aligned}$$
$$\begin{aligned}\Delta(c^{\underline x})&=c^{\underline{x+1}}-c^{\underline x}\\&=(c-1)c^{\underline x}-xc^{\underline x}.\end{aligned}$$
$$\begin{aligned}\sum_{k=1}^n\frac{(-2)^{\underline k}}k&=\sum_{1\leq k\lt n+1}(k+1)(-2)^{\underline{k-2}}\\&=\sum_{1\leq k\lt n+1}(k-2)(-2)^{\underline{k-2}}-(-3)(-2)^{\underline{k-2}}\\&=-(-2)^{\underline{k-2}}\Big|_{1}^{n+1}\\&=(-1)^nn!-1.\end{aligned}$$
From the second line to the third line, the interchange of summation is not justifiable. That’s because the terms in
$$\sum_{k\geq1}\sum_{j\geq1}\left(\frac{k}j[j=k+1]-\frac{j}k[j=k-1]\right)$$
do not converge absolutely.
