using pmi prove that 3.2^2+3^2.2^3.......+3^n.2^(n+1)=(12/5)(6^n-1)

To prove the equation \(3.2^2 + 3^2.2^3 + \ldots + 3^n.2^{n+1} = \frac{12}{5}(6^n - 1)\) using the principle of mathematical induction (PMI), we will follow a structured approach. This involves two main steps: establishing a base case and then proving that if the statement holds for an arbitrary case \(n = k\), it also holds for \(n = k + 1\).

Base Case

Let's start with the base case where \(n = 1\). We need to evaluate the left-hand side:

For \(n = 1\):

• Left-hand side: \(3^1 \cdot 2^{1 + 1} = 3 \cdot 2^2 = 3 \cdot 4 = 12\)
• Right-hand side: \(\frac{12}{5}(6^1 - 1) = \frac{12}{5}(6 - 1) = \frac{12}{5} \cdot 5 = 12\)

Since both sides equal 12, the base case holds true.

Inductive Step

Next, we assume that the statement is true for \(n = k\), which means:

Assumption: \(3.2^2 + 3^2.2^3 + \ldots + 3^k.2^{k + 1} = \frac{12}{5}(6^k - 1)\)

Now, we need to prove that it holds for \(n = k + 1\):

We consider the left-hand side for \(n = k + 1\):

• Left-hand side: \(3.2^2 + 3^2.2^3 + \ldots + 3^k.2^{k + 1} + 3^{k + 1}.2^{(k + 1) + 1}\)
• Using our assumption, this becomes: \(\frac{12}{5}(6^k - 1) + 3^{k + 1}.2^{k + 2}\)

Now, we need to simplify this expression:

We know that \(6^k = 3^k \cdot 2^k\). Thus, we can express \(3^{k + 1}\) as \(3 \cdot 3^k\) and \(2^{k + 2}\) as \(2^2 \cdot 2^k\). This gives us:

\(3^{k + 1}.2^{k + 2} = 3 \cdot 3^k \cdot 4 \cdot 2^k = 12 \cdot 3^k \cdot 2^k = 12 \cdot 6^k\)

Now substituting back into our equation:

\(\frac{12}{5}(6^k - 1) + 12 \cdot 6^k\)

Combining the terms gives us:

\(\frac{12}{5}(6^k - 1) + \frac{60}{5} \cdot 6^k = \frac{12 + 60}{5} \cdot 6^k - \frac{12}{5}\)

This simplifies to:

\(\frac{72}{5} \cdot 6^k - \frac{12}{5} = \frac{72 \cdot 6^k - 12}{5}\)

Factoring out \(\frac{12}{5}\) gives us:

\(\frac{12}{5}(6^{k + 1} - 1)\)

Conclusion of the Inductive Step

Thus, we have shown that if the statement holds for \(n = k\), it also holds for \(n = k + 1\). Since both the base case and the inductive step have been verified, by the principle of mathematical induction, we conclude that:

For all integers \(n \geq 1\), the equation \(3.2^2 + 3^2.2^3 + \ldots + 3^n.2^{n + 1} = \frac{12}{5}(6^n - 1)\) is true.

Approved