$5 \times 1-4=\frac{ 1\left( 5 \times 1-3 \right) }{ 2 }$
Check the equality$\textnormal{True}$
Write the inductive hypothesis by assuming that the statement $1+6+11+16+…+\left( 5n-4 \right)=\frac{ n \times \left( 5n-3 \right) }{ 2 }$ is true for $n=k$$1+6+11+16+…+\left( 5k-4 \right)=\frac{ k \times \left( 5k-3 \right) }{ 2 }$
To perform the inductive step and prove that the statement holds true for $n=k+1$, add ($k+1$)th term to both sides of the equality$1+6+11+16+…+\left( 5k-4 \right)+5\left( k+1 \right)-4=\frac{ k \times \left( 5k-3 \right) }{ 2 }+5\left( k+1 \right)-4$
Transform the right side of the equality$1+6+11+16+…+\left( 5k-4 \right)+5\left( k+1 \right)-4=\frac{ \left( k+1 \right) \times \left( 5\left( k+1 \right)-3 \right) }{ 2 }$
The formula $1+6+11+16+…+\left( 5n-4 \right)=\frac{ n \times \left( 5n-3 \right) }{ 2 }$ holds for $n=k+1$ whenever it holds for $n=k$, so by the principle of mathematical induction, the statement is true for all positive integers $n$$\textnormal{True}$