Inductive proofs are fun.

Binomial coefficients

These series identities are based on exercises from the first chapter of Statistical Inference¹

Alternating sum

For $n\ge 2$, $$\sum _{k=0}^n{(-1)}^k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$$

For $n=2$, ${\sum }_{k=0}^2{(-1)}^k\left(\genfrac{}{}{0ex}{}{2}{k}\right)=1-2+1=0$.

Induction hypothesis (IH): assume that this holds for $n\ge 2$. We want to show that $$\sum _{k=0}^{n+1}{(-1)}^k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$$

follows. Use Pascal's rule:

$$\begin{array}{ccc}\hfill \left(\genfrac{}{}{0ex}{}{n}{k}\right)& =\frac{n!}{k!(n-k)!}\hfill & \text{(1)}\\ \hfill & =\frac{(n-1)!(k+n-k)}{k!(n-k!)}\hfill & \text{(2)}\\ \hfill & =(n-1)!(\frac{k}{k!(n-k)!}+\frac{n-k}{k!(n-k)!})\hfill & \text{(3)}\\ \hfill & =\frac{(n-1)!}{(k-1)!(n-1-k+1)!}+\frac{(n-1)!}{k!(n-1-k)!}\hfill & \text{(4)}\\ \hfill & =\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\hfill & \text{(5)}\end{array}$$

Write $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ as $a_k$. Then,

$$\begin{array}{ccc}\hfill & \sum _{k=0}^{n+1}(-1{)}^k(\genfrac{}{}{0ex}{}{n+1}{k})\hfill & \text{(1)}\\ \hfill & =1+\sum _{k=1}^n(-1{)}^k\left(\right(\genfrac{}{}{0ex}{}{n}{k-1})+(\genfrac{}{}{0ex}{}{n}{k}\left)\right)+(-1{)}^{n+1}·1\hfill & \text{(2)}\\ \hfill & =1+\sum _{k=1}^n(-1{)}^k{a}_k+\sum _{k=1}^n(-1{)}^k{a}_{k-1}+(-1{)}^{n+1}·1\hfill & \text{(3)}\\ \hfill & =\sum _{k=0}^n(-1{)}^k{a}_k+\sum _{k=1}^n(-1{)}^k{a}_{k-1}+(-1{)}^{n+1}·a_n\hfill & \text{(4)}\\ \hfill & =0-\sum _{k=0}^{n-1}(-1{)}^k{a}_k-(-1{)}^n·a_n\hfill & \text{(5)}\\ \hfill & =-({\sum }_{k=0}^{n-1}(-1{)}^k{a}_k+(-1{)}^n·a_n)\hfill & \text{(6)}\\ \hfill & =-({\sum }_{k=0}^n(-1{)}^k{a}_k)\hfill & \text{(7)}\\ \hfill & =0\hfill & \text{(8)}\end{array}$$