By use of mathematical induction proove that a) [ 3+6+12+...+3 n=(3 n(n+1))/(2) ]

To prove the given statement using mathematical induction, we will follow the standard steps of induction.<br /><br />**Base Case:**<br />Let's start by checking the base case, which is when \( n = 1 \).<br /><br />Substituting \( n = 1 \) into the given equation, we have:<br />\[ 3 + 6 + 12 + \cdots + 3 \cdot 1 = \frac{3 \cdot 1 \cdot (1 + 1)}{2} \]<br /><br />Simplifying the left-hand side, we get:<br />\[ 3 = \frac{3 \cdot 1 \cdot 2}{2} = 3 \]<br /><br />So, the base case holds true.<br /><br />**Inductive Step:**<br />Now, let's assume that the statement is true for some arbitrary positive integer \( k \), i.e., we assume that:<br />\[ 3 + 6 + 12 + \cdots + 3k = \frac{3k(k+1)}{2} \]<br /><br />We need to prove that the statement is true for \( k + 1 \), i.e., we need to show that:<br />\[ 3 + 6 + 12 + \cdots + 3k + 3(k+1) = \frac{3(k+1)(k+2)}{2} \]<br /><br />Using the inductive hypothesis, we can rewrite the left-hand side as:<br />\[ \frac{3k(k+1)}{2} + 3(k+1) \]<br /><br />Now, let's simplify the right-hand side:<br />\[ \frac{3k(k+1)}{2} + 3(k+1) = \frac{3k(k+1) + 6(k+1)}{2} = \frac{3(k+1)(k+2)}{2} \]<br /><br />Thus, we have shown that the statement holds true for \( k + 1 \) if it holds true for \( k \).<br /><br />By the principle of mathematical induction, we can conclude that the given statement is true for all positive integers \( n \).