Question

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

 

Answer 1

Let P(n) be the given statement.
Now, 

\[P(n): a + (a + d) + (a + 2d) + . . . + (a + (n - 1)d) = \frac{n}{2}\left[ 2a + (n - 1)d \right]\]
\[\text{ Step1} : \]
\[P(1) = a = \frac{1}{2}(2a + (1 - 1)d)\]
\[\text{ Hence, P(1) is true .}  \]
\[\text{ Step 2: } \]
\[\text{ Suppose P(m) is true .}  \]
\[\text{ Then,}  \]
\[a + (a + d) + . . . + (a + (m - 1)d) = \frac{m}{2}\left[ 2a + (m - 1)d \right]\]
\[\text{ We have to show that P(m + 1) is true whenever P(m) is true }  . \]
\[\text{ That is,}  \]
\[a + (a + d) + . . . + (a + md) = \frac{(m + 1)}{2}\left[ 2a + md \right]\]
\[\text{ We know that P(m) is true}  . \]
\[\text{ Thus, we have:}  \]
\[a + (a + d) + . . . + (a + (m - 1)d) = \frac{m}{2}\left[ 2a + (m - 1)d \right]\]
\[ \Rightarrow a + (a + d) + . . . + (a + (m - 1)d) + (a + md) = \frac{m}{2}\left[ 2a + (m - 1)d \right] + (a + md) \left[ \text{ Adding}  (a + md) \text{ to both sides } \right]\]
\[ \Rightarrow P(m + 1) = \frac{1}{2}\left[ 2am + m^2 d - md + 2a + 2md \right]\]
\[ \Rightarrow P(m + 1) = \frac{1}{2}\left[ 2a(m + 1) + md(m + 1) \right]\]
\[ = \frac{1}{2}(m + 1)(2a + md)\]
\[\text{ Thus, P(m + 1) is true . }  \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n }  \in N .\]