# Evaluate: 1+6+11+16+\dots+(5n-4)=(n * (5n-3))/(2)

## Expression: $1+6+11+16+…+\left( 5n-4 \right)=\frac{ n \times \left( 5n-3 \right) }{ 2 }$

To perform the anchor step, write the equality for $n=1$

$5 \times 1-4=\frac{ 1\left( 5 \times 1-3 \right) }{ 2 }$

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 $n$

$\textnormal{True}$

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