Question

prove that 7 is factor of 2³ⁿ - 1 for all natural numbers n , n € R ​

Answer 1

Answer:

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Answer 2

To ProvE :-

• To prove that , $\sf 2^{3n}-1$ is a factor of all real numbers , $\sf n\in \mathbb{R}$ .

ProoF :-

We will prove this by the process of mathematical induction . The process of mathematical induction ihelps us to prove some formulae which cannot be proved easily by direct methods .

$\red{\bigstar}\underline{\textsf{ Principal of Mathematical Induction :- }}$

$\textsf{ If P(n) is a statement such that , }$

• P(n) is true for n = 1
• P(n) is true for n = k + 1 , when it's also true for n = k , where $\sf k \in \mathbb{N}$

Then we can say that by the principal of mathematical induction , the statement is true for all Real Numbers .

$\rule{200}2$

$\sf\to Let \ \pink{P(n)} : 7\ is \ a\ factor\ of\ 2^{3n}-1\ for \ n \ \in \mathbb{R} .$

$\red{\bigstar}\textsf{\textbf{ Step 1 :- }\underline{ Put n = 1 :- }}$

$\sf \to 2^{3(1)} -1 \\\\\sf \to 2^3-1 \\\\\sf\to 8 - 1 \\\\\sf\to \blue{ 7 } .$

Here 7 is a factor of itself . So we can say that the statement is true for n = 1 .

$\sf\dashrightarrow\purple{ P(n) \ is \ true\ for \ n = 1 .}$

$\rule{200}2$

$\red{\bigstar}\textsf{\textbf{ Step 2:- }\underline{ Let us take that it is true for n = k :- }}$

$\sf\to 7\ is\ a\ factor\ of \ \red{2^{3k}-1}$

$\sf\to Let\ then ,\ 2^{3k}-1 = 7A \qquad\bigg\lgroup \red{\tt Where \ A \ \in \mathbb{N} .}\bigg\rgroup$

$\sf\to\pink{ 2^{3k} = 7A + 1}$

Now do the same thing by substituting n = (k+1) in the given statement .

• So that :-

$\sf\to 2^{3(k+1)} -1$

$\sf\to 2^{3k + 3 }$

$\sf \to 2^{3k}.2^{3}-1 \qquad\bigg\lgroup \red{\tt Using \ a^m\times a^n = a^{m+n}}\bigg\rgroup \\\\\sf\to (7A+1)8 - 1\qquad\bigg\lgroup \red{\tt From \ above }\bigg\rgroup\\\\\sf\to 56A+7 \\\\\sf\to \pink{7(8A+1)}.$

This implies that ,

$\implies\tt 2^{3(k+1)} = 7(8A+1)$

Therefore 7 is a factor of $\sf 2^{3(k+1)}-1$

$\sf\dashrightarrow\purple{ P(n) \ is \ true\ for \ n = (k+1) .}$

$\rule{200}2$

Therefore by the Principal of Mathematical induction we can say that ,

$\implies\pink{\sf 2^{3n}-1 }$ is a factor of all real numbers ,$\sf n\in \mathbb{R}$ .

$\rule{200}2$