Name: Definition of Axiom, Conjecture and Theorem
Uploaded: 2019-07-19
Duration: 2 min 6 s
Description: The origins of mathematics have a lot in common with philosophy. So, it's no surprise that we study axioms, conjectures, and theorems in mathematics. Learn m...

Question

3 \cdot 2^{2} + 3^{2} 2^{3} + 3^{3} 2^{4} + \dots \dots + 3^{n} 2^{n + 1} = \frac{12}{5} (6^{n} - 1)

Accepted Answer

Let

$P (n) : 3 \cdot 2^{2} + 3^{2} 2^{3} + 3^{3} 2^{4} + \dots \dots + 3^{n} 2^{n + 1} = \frac{12}{5} (6^{n} - 1)$

Now,

$P (1) \equiv 3 \times 2^{2} = \frac{12}{5} (6 - 1)$

Hence, $P (1)$ is true.

Let $P (k)$ is true, then

$3 \cdot 2^{2} + 3^{2} 2^{3} + 3^{3} 2^{4} + \dots \dots + 3^{k} \cdot 2^{k + 1} = \frac{12}{5} (6^{k} - 1)$

Now,

$P (k + 1) : 3 \cdot 2^{2} + 3^{2} 2^{3} + 3^{3} 2^{4} + \dots \dots + 3^{k} \cdot 2^{k + 1} + 3^{k + 1} \cdot 2^{k + 2}$

$\Rightarrow P (k + 1) : [\frac{12}{5} (6^{k} - 1)] + 3^{k + 1} \cdot 2^{k + 2}$

$\Rightarrow P (k + 1) : [\frac{12}{5} (6^{k} - 1)] + 12 \cdot 3^{k} \cdot 2^{k}$

$\Rightarrow P (k + 1) : [\frac{12}{5} (6^{k} - 1)] + 12 \cdot 6^{k}$

$\Rightarrow P (k + 1) : [(\frac{12}{5} + 12) 6^{k} - \frac{12}{5}]$

$\Rightarrow P (k + 1) : [(\frac{72}{5}) 6^{k} - \frac{12}{5}]$

$\Rightarrow P (k + 1) : [(\frac{72}{5 \times 6}) 6^{k + 1} - \frac{12}{5}]$

$\Rightarrow P (k + 1) : [(\frac{12}{5}) 6^{k + 1} - \frac{12}{5}]$

$\Rightarrow P (k + 1) : \frac{12}{5} (6^{k + 1} - 1)$

Hence, $P (k + 1)$ is also true when $P (k)$ is true. So, $P (n)$ is true for all $n \in N$ .

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