Question

by using factor theorem
Prove that 49¹¹¹¹¹ - 1 is divisible by 48.​

Answer 1

Answer:

your proof is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: prove \: that : \\ 49 {}^{n} - 1 \: \: is \: divisible \: \: by \: 48 \\ \\ we \: would \: prove \: this \: by \\ principle \: of \: mathematical \: \\ induction \: \: method \\ \\ Step \: I : \\ let \: here \: then\\ P(n) = 49 {}^{n} \: - 1 \: \: be \: divisible \: \: by \: 48 \\ put \: n = 1 \\ \\ 49 {}^{n} - 1 = 49 {}^{1} - 1 \\ = 49 - 1 \\ = 48 \\ \\ which \: is \: divisible \: by \: 48 \\ hence \: here \: \: P(1) \: istrue$

$Step \: II : \\ assume \: \: that \: P(k) \: is \: true \: \: ie \: \\ P(k) = 49 {}^{k } - 1 = 48a \\ \\ 49 {}^{k} = 48a + 1 \: \: \: \: \: (1)$

$Step \: III : \\ to \: prove \: that \: \: P(k + 1) \: \: is \: true \\ let \: then \: here \\ P(k + 1) = 49 {}^{k + 1} - 1 = 48b \\ \\ =( 49 {}^{k} . \: 49) - 1 \\ \\ = [(48a + 1) \times 49 \: ] - 1 \\ \\ = 49.48a + 49 - 1 \\ \\ = 49.48a + 48 \\ \\ = 48(49a + 1) \\ \\ which \: is \: a \: multiple \: and \: divisible \\ of \: 48 \: and \: by \: 48 \: \: respectively \\ \\ hence \: here \\ P(k + 1) \: \: is \: true \: \: for \: all \\ n∈N$

$Step \: IV : \\ hence \: by \: principle \: of \: \\ mathematical \: induction \: \\ we \: prove \: that \\ P(n) \: is \: true \\ \\ ∴ \: 49 {}^{n} - 1 \: \: is \: divisible \: by \: 48 \\ for \: all \: n∈N$