Question

In using induction to prove that 7^(2n + 1) + 3^(2n +1) is divisible by 10, what statement can be represented by Sk + 1 ?
7^(2n + 1) + 3^(2n + 1) = 10p where p is an integer
7^(2n + 3) + 3^(2n + 3) = 10p where p is an integer
7^(2n + 1) + 3^(2n + 1) = 10
7^(2n + 3) + 3^(2n + 3) = 10

Answer 1

Do you mean S(k+1)?

Answer 2

yes sorry about that

Answer 3

Your statement 7^(2n + 1) + 3^(2n +1) is S(n), so to get S(k+1), replace n with (k+1): \[\large 7^{2(k+1) + 1} + 3^{2(k+1) +1} =7^{2k+2+1}+3^{2k+2+1} =7^{2k+3}+3^{2k+3}\]
But I think they just reverted back to using the variable n instead, so just replace your k's with n's.

Now you are left with 2 possible answers.

Answer 4

thank so its b or d

Answer 5

yes. Now the key here is that you want your expression to be divisible by 10. That means that when you divide that long expression by 10, you get an integer, i.e.:
\[\large \frac{7^{2k+3}+3^{2k+3}}{10}=\text{integer} \]

Just think with regular numbers... You know 90 is divisible by 10 since 90/10 = 9, an integer. 91 is not divisible by 10 since 91/10 = 9.1, not an integer.

Answer 6

so its D. Thanks great help

Answer 7

actually it's not 10 :S

Answer 8

ohh its b

Answer 9

\[ 7^{2n + 1} + 3^{2n + 1} = 10p\]
divide both sides by 10:
\[\frac{7^{2n + 1} + 3^{2n + 1}}{10} =p\] and p is the integer

Answer 10

thanks i get it now