Question

Prove by the method of mathematical induction that 

a + (a + d) + (a + 2d) + ... + (a + (n – 1) d) = \(\frac{n}{2}\) {2a + (n – 1) d} for all n∈ N, where a, d∈R.

Answer 1

Let T(n) be the statement 

a + (a + d) + (a + 2d) + ... + (a + (n – 1)d) = \(\frac{n}{2}\) [2a + (n – 1)d]

Basic Step: 

For n = 1, LHS = a,  RHS = \(\frac{1}{2}\)[2a] = a 

⇒ LHS = RHS ⇒ T(1) is true.

Induction Step: 

Let T(k) hold true, i.e., 

a + (a + d) + (a + 2d) + ... + (a +(k – 1)d) = \(\frac{k}{2}\) [2a +(k – 1)d] 

Now to show that T (k + 1) holds true, we add the (k + 1)^th term, i.e., a + {(k + 1) – 1} d = a + kd to both the sides of T(k), i.e., 

a + (a + d) + (a + 2d) + ... + (a + (k – 1) d) + (a + kd)

= \(\frac{k}{2}\) [2a+(k-1)d] +(a+kd) = ak + \(\frac{k(k-1)d}{2}\) +a + kd

= a (k + 1) + \(\frac{1}{2}\) {k(k – 1)d + 2kd} = (k + 1) a + \(\frac{1}{2}\) {k²d + kd} 

= (k + 1) a + \(\frac{1}{2}\)k(k + 1)d = \(\frac{(k+1)}{2}\) + [2a + {(k + 1) – 1}d]. 

Thus, T (k + 1) is true, whenever T(k) is true.