Prove by induction that f(n) = 2^(k + 2) + 3^(3k + 1) is divisible by 7 for all positive n.

First we establish our base case: f(0) = 22 + 31 = 4 + 3 = 7, so clearly f(0) is divisible by 7.Now. by the inductive hypothesis. we assume that f(k) is divisible by 7, and attempt to show that this implies f(k+1) is also divisible by 7.f(k + 1) = 2k + 3 + 32(k + 1) + 1 = 2k + 3 + 32k + 3 = 2 * 2k + 2 + 9 * 32k + 1 So f(k + 1) mod 7 === 2 * 2k + 2 + 2 * 32k + 1 (since 9 mod 7 === 2). So f(k + 1) mod 7 === 2 * (2k + 2 + 32k + 1) = 2 * f(k).These for f(k + 1) mod 7 === 0, hence f(k + 1) is divisible by 7 if f(k) is divisible by 7, hence f(n) is divisibile by 7 for all n >= 0.

WP
Answered by William P. Further Mathematics tutor

2357 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Use de Moivre's theorem to calculate an expression for sin(5x) in terms of sin(x) only.


Evaluate (1 + i)^12


It is given that z = 3i(7-i)(i+1). Show that z can be written in the form 24i - k. State the integer k.


Given that the equation x^2 - 2x + 2 = 0 has roots A and B, find the values A + B, and A * B.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences