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

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Solution

Let P(n) be the given statement.
Now,
$P (n) : a + (a + d) + (a + 2 d) + . . . + (a + (n - 1) d) = \frac{n}{2} [2 a + (n - 1) d] Step 1 : P (1) = a = \frac{1}{2} (2 a + (1 - 1) d) Hence, P (1) is true . Step 2 : Suppose P (m) is true . Then, a + (a + d) + . . . + (a + (m - 1) d) = \frac{m}{2} [2 a + (m - 1) d] We h a v e t o show that P (m + 1) is true whenever P (m) is true . T h a t i s, a + (a + d) + . . . + (a + m d) = \frac{(m + 1)}{2} [2 a + m d] W e know that P (m) is true . T h u s, w e h a v e : a + (a + d) + . . . + (a + (m - 1) d) = \frac{m}{2} [2 a + (m - 1) d] \Rightarrow a + (a + d) + . . . + (a + (m - 1) d) + (a + m d) = \frac{m}{2} [2 a + (m - 1) d] + (a + m d) [Adding (a + m d) to both sides] \Rightarrow P (m + 1) = \frac{1}{2} [2 a m + m^{2} d - m d + 2 a + 2 m d] \Rightarrow P (m + 1) = \frac{1}{2} [2 a (m + 1) + m d (m + 1)] = \frac{1}{2} (m + 1) (2 a + m d) T h u s, P (m + 1) is true . B y t h e p rinciple of m athematical i nduction, P (n) is true for all n \in N .$

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