Question

Use mathematical induction to prove this formula for the sum of the terms of an arithmetic progression.

a+(a+d) + ... + (a+nd) = (n+1)(2a+nd) / 2

Short answer

It is proved that a+(a+d) + ... + (a+nd) = (n+1)(2a+nd) / 2

Step-by-step solution

Principle of Mathematical Induction

Consider the propositional function \(P\left( n \right)\). Consider two actions to prove that\(P\left( n \right)\) evaluates to accurate for all set of positive integers\(n\).

Consider the first basic step is to confirm that\(P\left( 1 \right)\)true.

Consider the inductive step is to demonstrate that for any positive integer k the conditional statement \(P\left( k \right) \to P\left( {k + 1} \right)\)is true.

Prove the basis step

Let \(P\left( n \right)\):\(a + \left( {a + d} \right) + ... + \left( {a + nd} \right) = \frac{{\left( {n + 1} \right)\left( {2a + nd} \right)}}{2}\)

In the basis step, one need to prove that\(P\left( 1 \right)\)is true.

For finding statement\(P\left( 1 \right)\)substituting\(1\)for\(n\)in the statement.

\(\begin{array}{c}a + \left( {a + \left( 1 \right)d} \right) = \left( {1 + 1} \right){{\left( {2a + \left( 1 \right)d} \right)} \mathord{\left/

{\vphantom {{\left( {2a + \left( 1 \right)d} \right)} 2}} \right.

\kern-\nulldelimiterspace} 2}\\2a + d = 2a + d\end{array}\)

From the above, we can see that the statement \(P\left( 1 \right)\) is true this is also known as the basis step of the proof.

Prove the Inductive step

In the inductive step, we need to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.

\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

In the inductive hypothesis, we assume that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\).

Solve as:

\(a + \left( {a + d} \right) + ... + \left( {a + kd} \right) = \frac{{\left( {k + 1} \right)\left( {2a + kd} \right)}}{2}\)

To show that\(P\left( {k + 1} \right)\)is also true replace\(k\)with\(k + 1\)in the statement.

\(a + \left( {a + d} \right) + ... + \left( {a + \left( {k + 1} \right)d} \right) = \frac{{\left( {k + 1 + 1} \right)\left( {2a + \left( {k + 1} \right)d} \right)}}{2}\)

Now, add\(\left( {a + \left( {k + 1} \right)d} \right)\)in both sides of the equation (i) or inductive hypothesis.

\(\begin{aligned}{c}a + \left( {a + d} \right) + ... + \left( {a + kd} \right) + \left( {a + \left( {k + 1} \right)d} \right) &= \frac{{\left( {k + 1} \right)\left( {2a + kd} \right)}}{2} + \left( {a + \left( {k + 1} \right)d} \right)\\ &= \frac{{\left( {k + 1} \right)\left( {2a + kd} \right) + 2\left( {a + \left( {k + 1} \right)d} \right)}}{2}\\ &= \frac{{2a\left( {k + 1 + 1} \right) + d({k^2} + k + 2k + 2)}}{2}\\ &= \frac{{\left( {k + 1 + 1} \right)\left( {2a + d(k + 1)} \right)}}{2}\end{aligned}\)

From the above, it is seen that\(P\left( {k + 1} \right)\)is also true.

Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This

completes the inductive step.

Hence it is proved that

\(a + \left( {a + d} \right) + ... + \left( {a + nd} \right) = \left( {n + 1} \right){{\left( {2a + nd} \right)} \mathord{\left/

{\vphantom {{\left( {2a + nd} \right)} 2}} \right.

\kern-\nulldelimiterspace} 2}\).