Mathematical Induction - divisibility (1 Viewer)

[j1mmy_]

I am stuck of step 3.

Here's what I've done so far.

Step 1: Prove true for n = 1





Step 2: Assume n = k


Step 3: Prove true for n = k+1



and now I'm stuck after this.

 

[HeroicPandas]

5.5^k + 22. 11^k = 15M + 12 .11^k = 3(5M + 4.11^k) = 3Q

 

[RivalryofTroll]

I am stuck of step 3.

Here's what I've done so far.

Step 1: Prove true for n = 1





Step 2: Assume n = k


Step 3: Prove true for n = k+1



and now I'm stuck after this.

5^k = 3M - 2(11)^k [from step 2]

So continuing....
= 5(3M - 2(11)^k) + (22.(11)^k)
= 15M - 10(11)^k + 22(11)^k
= 15M + 12(11)^k
= 3(5M + 4(11)^k)
= 3Q (Where Q is an integer)

Thus, true for n=k+1

 

[j1mmy_]

= 3(5M + 4(11)^k) ...(1)
= 3Q (Where Q is an integer) ...(2)

Are we allowed to stop at (1), or do we have to do (2). "Q" is pretty much letting it to be the stuff in the brackets right? (ie. 5M + 4(11)^k)

 

[RivalryofTroll]

Are we allowed to stop at (1), or do we have to do (2). "Q" is pretty much letting it to be the stuff in the brackets right? (ie. 5M + 4(11)^k)

Yeah i guess you can stop at (1) and say, which is divisible by 3.

Yeah Q is just an integer.

 

[j1mmy_]

Yeah i guess you can stop at (1) and say, which is divisible by 3.

Yeah Q is just an integer.

Alright thanks!