Prove that $4+7+10+...+\left(3n+1\right)=\frac{n\left(3n+5\right)}{2}$ for all positive integers n.

Given, to prove that the statement $4+7+10+...+\left(3n+1\right)=\frac{n\left(3n+5\right)}{2}$ is true for all positive integers n.

When n=1, the statement is written as:

$\begin{array}{l}4+7+10+\mathrm{...}+\left(3n+1\right)=\frac{n\left(3n+5\right)}{2}\\ 4=\frac{\left(1\right)\left(3\left(1\right)+5\right)}{2}\\ 4=\frac{3+5}{2}\\ 4=\frac{8}{2}\\ 4=4\end{array}$

Hence the statement is true for n = 1.

Let the statement be true for n = k. Hence,

$4+7+10+...+\left(3k+1\right)=\frac{k\left(3k+5\right)}{2}$

Checking the statement when n = k + 1:

$\begin{array}{l}4+7+10+\mathrm{...}+\left(3n+1\right)=\frac{n\left(3n+5\right)}{2}\\ 4+7+10+\mathrm{...}+\left(3\left(k+1\right)+1\right)=\frac{\left(k+1\right)\left(3\left(k+1\right)+5\right)}{2}\\ 4+7+10+\mathrm{...}+\left(3k+1\right)+\left(3k+4\right)=\frac{\left(k+1\right)\left(3k+8\right)}{2}\end{array}$

Plugging the value from the previous assumptions:

$\begin{array}{l}\frac{k\left(3k+5\right)}{2}+\left(3k+4\right)=\frac{\left(k+1\right)\left(3k+8\right)}{2}\\ \frac{k\left(3k+5\right)+2\left(3k+4\right)}{2}=\frac{\left(k+1\right)\left(3k+8\right)}{2}\\ \frac{3k^2+5k+6k+8}{2}=\frac{\left(k+1\right)\left(3k+8\right)}{2}\\ \frac{3k^2+3k+8k+8}{2}=\frac{\left(k+1\right)\left(3k+8\right)}{2}\\ \frac{3k\left(k+1\right)+8\left(k+1\right)}{2}=\frac{\left(k+1\right)\left(3k+8\right)}{2}\\ \frac{\left(k+1\right)\left(3k+8\right)}{2}=\frac{\left(k+1\right)\left(3k+8\right)}{2}\end{array}$

Hence, the statement is true for n = k +1.

Hence by mathematical induction, the given statement is true for all positive integers n.

Hence,by mathematical induction,$4+7+10+...+\left(3n+1\right)=\frac{n\left(3n+5\right)}{2}$ is true for all positive integers n.