- #1

- 1

- 0

so this is what i could do so far:

for n=1

3^(2*1-1)+1=4 which is divisible by 4

assume truth for n=k

(3^(2k-1))+1 is divisible by 4

and i know that next i have to prove for n=k+1 but i really have no idea what to do witht that.

please help

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- Thread starter yelena0000
- Start date

- #1

- 1

- 0

so this is what i could do so far:

for n=1

3^(2*1-1)+1=4 which is divisible by 4

assume truth for n=k

(3^(2k-1))+1 is divisible by 4

and i know that next i have to prove for n=k+1 but i really have no idea what to do witht that.

please help

- #2

Hitman2-2

and i know that next i have to prove for n=k+1 but i really have no idea what to do with that.

If you substitute

[tex]

3^{2(k+1) - 1} + 1 = 3^{2k-1} 3^2 + 1

[/tex]

By assumption, [tex] 4 | 3^{2k-1} + 1 [/tex] so try to re-write [tex] 3^{2k-1} 3^2 + 1 [/tex] in a form that has a factor of [tex] 3^{2k-1} + 1 [/tex].

- #3

statdad

Homework Helper

- 1,495

- 36

The general term is

[tex]

a_n = 3^{2n-1}+1

[/tex]

and you've shown that $a_1$ is divisible by four, and you've assumed the same for $a_k$ for some $k \ge 1$.

Look at [tex] a_{k+1} [/tex].

[tex]

3^{2(k+1)-1} +1 = 3^{2k+2-1} + 1 = 3^{2k -1} 3^2 + 1

[/tex]

The goal is to show that this is also divisible by four - the fact that [tex] a_k [/tex] is divisible by four will play a role in this.

[tex]

a_n = 3^{2n-1}+1

[/tex]

and you've shown that $a_1$ is divisible by four, and you've assumed the same for $a_k$ for some $k \ge 1$.

Look at [tex] a_{k+1} [/tex].

[tex]

3^{2(k+1)-1} +1 = 3^{2k+2-1} + 1 = 3^{2k -1} 3^2 + 1

[/tex]

The goal is to show that this is also divisible by four - the fact that [tex] a_k [/tex] is divisible by four will play a role in this.

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