proof by induction

[shelly89]

prove by induction that \(\displaystyle 11 | 5^{2n}-3^{n} \)
let f(n) = \(\displaystyle 5^{2n}-3^{n} /tex]



f(1) = \(\displaystyle 5^{2} - 3 = 22 \) which is divisible by 11

assume it is true for n =k

f(k) = \(\displaystyle 5^{2k}-3^{k} \)

induction step


f(k+1) = \(\displaystyle 5^{2{k+1}}-3^{k+1} \) = \(\displaystyle 5^{2}(5^{2k})- 3(3^{k}) \)

f(k+1) - f(k) = \(\displaystyle 25(5^{2k}) - 3(3^{k}) - f(k) \)

f(k+1) - f(k) = \(\displaystyle 24(5^{2k}) - 2(3^{k}) \)

Now I am not sure where to go from here,

I have tried to split the numbers.....

f(k+1) - f(k) = \(\displaystyle 22(5^{2k}) + 2(5^{2k} -2(3^{k})) \)

f(k+1) = \(\displaystyle f(k) +22(5^{2k}) + 2(f(k)) \)

f(k+1) = \(\displaystyle 22(5^{2k}) + 3(f(k)) \)

Have I proved it ?\)

 

[HallsofIvy]

prove by induction that \(\displaystyle 11 | 5^{2n}-3^{n} \)
let f(n) = \(\displaystyle 5^{2n}-3^{n} /tex]



f(1) = \(\displaystyle 5^{2} - 3 = 22 \) which is divisible by 11

assume it is true for n =k

f(k) = \(\displaystyle 5^{2k}-3^{k} \)

induction step


f(k+1) = \(\displaystyle 5^{2{k+1}}-3^{k+1} \) = \(\displaystyle 5^{2}(5^{2k})- 3(3^{k}) \)\)

\(\displaystyle
You have a typo here, with 2{k+ 1} where you want 5(k+1).\(\displaystyle 5^{2(k+1)}\), not \(\displaystyle 5^{2{k+1}}\).

f(k+1) - f(k) = \(\displaystyle 25(5^{2k}) - 3(3^{k}) - f(k) \)

f(k+1) - f(k) = \(\displaystyle 24(5^{2k}) - 2(3^{k}) \)

Now I am not sure where to go from here,

I have tried to split the numbers.....

f(k+1) - f(k) = \(\displaystyle 22(5^{2k}) + 2(5^{2k} -2(3^{k})) \)

f(k+1) = \(\displaystyle f(k) +22(5^{2k}) + 2(f(k)) \)

f(k+1) = \(\displaystyle 22(5^{2k}) + 3(f(k)) \)

Have I proved it ?

Almost. You began with the assumtion that f(k) was a multiple of 11: f(k)= 11n for some n so \(\displaystyle f(k+1)= 22(5^{2k}+ 3(11n)= 11(2(5^k)+ 3n)\)\)