#1**+1 **

\(P_1 : 7(7^1 - 1) = 7\cdot 6 = 42 \text{ is divisible by }42 \text{ so }P_1 \text{ is True}\\ \text{assume }P_n \text{ is True, and prove }P_{n+1} \text{ is True}\\ 7(7^{n+1}-1) = 7(7^n\cdot 7 - 1) = \\ 7(7^n\cdot 7-7+6) = 7\cdot 7(7^n - 1) + 42 \\ \text{we assume }7(7^n -1) \text{ is divisible by 42 i.e. = }42k \text{ for some }k \in \mathbb{N}\\ 7\cdot 7(7^n - 1) + 42 = 7\cdot 42k+42 = 42(7k+1) \\ \text{thus }7(7^{n+1}-1) \text{ is divisible by }42 \text{ and }P_{n+1} \text{ is True. QED}\)

Rom
Nov 27, 2018

#2**+2 **

Show that is is true for n = 1

7 (7^{1} - 1) = 7 (6) = 42 so...true !!!!

Assume it is true for n = k

That is :

7(7^{k} - 1) is divisible by 42

Note that we can write this as

(7* 7^{k} - 7)

Prove it is true for n = k + 1

7(7^{k+ 1} - 1) =

7 ( 7 * 7^{k} - ( 7 - 6) ) =

7 ( [7 * 7k - 7 ] + 6 ) =

7 ( 7 * 7^{k} - 7) + 7*6

Note that the first term is divisible by 42 since ( 7 * 7^{k} - 7) was assumed to be divisible by 42

And 7*6 is clearly divisible by 42

So ... 7(7^{k+ 1} - 1) is divisible by 42

CPhill
Nov 27, 2018