Question

prove by induction that for all integral values of n
2^(n+2) + 3^(2n+1) is divisible by 7

Answer 1

in equation form
\[2^{n+2} + 3^{2n+1} \] is divisible by 7 for integral n

Answer 2

so this is very easy first calcule for n=0 and result that is true so for n=1 make calcule and result that is true after this suppose that for n=k is true and after this you need to prove that this will be true for n=k+1 too

Answer 3

those are all by proving by induction method !!!

good luck !

bye

Answer 4

yep, i understnd the induction method quite well, but proving the n=k+1 has had me going around in circles for about an hour!

Answer 5

i can see that its
\[2^{k+3} + 3^{2k+3}\]

Answer 6

its how to factor out the 7

Answer 7

proof for n=1:

2^3 + 3^3 = 8 + 27 = 35

Therefore for n=1 it is divisible by 7.

Now suppose \[2^{n+2} + 3^{2n + 1}\]
is divisible by 7.

Prove it for n+1:

\[2^{n + 1 + 2} + 3^{2(n+1) + 1}\]

\[=2^{n+3} + 3^{2n + 3}\]

Write it like this:

\[2 * 2^{n+2} + 3^2 * 3^{2n+2}\]

Factor out the 2:

\[2(2^{n+2} + 3^{2n + 1} ) + 3^2*3^{2n+1} - 2 * 3^{2n+1}\]

Factorize:

\[2(2^{n+2} + 3^{2n+1}) + 3^{2n+1} * (3^2 - 2)\]

\[2(2^{n+2} + 3^{2n+1}) + 3^{2n+1} * 7\]

Since:

\[2(2^{n+2} 3^{2n+1} )\]

is divisible by 7, since we said Suppose:

\[ ({} 2^{n+2} + 3^{2n + 1})\]

is divisible by 7.

now, the second part is also divisible by 7:

\[3^{2n+1} * 7\]

Because the term is multiplied by 7.

Therefore, because both terms are divisible by 7, this is true for n+1 and therefore all n >=1

This completes the proof.

Answer 8

It should be \[2(2^{n+2} + 3^{2n + 1})\]

and not

\[2(2^{n+2}3^{2n+1})\]

I made an error in the one line

Answer 9

briliant- thanks !

Answer 10

Shot