Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$

Question

I have tried to prove this through mathematical induction but I can't seem to prove that the proposition works for $k+1$.

Answer 1

Assuming that $7|(3^{2n-1} + 2^{n+1})$ then

$$3^{2n+1} + 2^{n+2}=3^2 \cdot 3^{2n-1} +2\cdot 2^{n+1}\equiv 2\cdot (3^{2n-1} + 2^{n+1}) \equiv 0\mod 7$$

which proves the induction step.

Answer 2

You can extract the factor $7$ as follows using

• $(\star)$: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$

Hence, $$3^{2n-1} + 2^{n+1}= 3\cdot 9^{n-1} + 4\cdot 2^{n-1}$$ $$= 3\cdot 9^{n-1} + (7-3)\cdot 2^{n-1} = 3(\underbrace{9^{n-1} - 2^{n-1}}_{\stackrel{ (\star)}{=}7\cdot m}) + 7\cdot 2^{n-1}$$

Answer 3

$3^{2n-1} \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &3 & 6 & 5 & 3 & 6 & \cdots \end{matrix}$

$2^{n+1} \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &4 & 1 & 2 & 4 & 1 & \cdots \end{matrix}$

Since $3+4 = 7$, $6+1=7$, $5+2 = 7$ we can see that $3^{2n-1}+2^{n+1} \equiv 0 \pmod{7}$

Answer 4

$(3^2)^{n-1}+2^{n+1}=$

$ (1/3)(7+2)^n +2 \cdot 2^n=$

$(1/3)(\sum_{k=0}^{n}\binom {n}{k} 7^{n-k}2^k+6\cdot 2^n)=$

$(1/3)(\sum_{k=0}^{n-1} \binom{n}{k} 7^{n-k} 2^k +7 \cdot 2^n)$;

All terms in the sum above have a factor $7$.

Answer 5

Hint: Write $$(3^2)^n\times3^{-1}+2^n\times 2$$ and $$3^2\equiv 2\mod n$$

Answer 6

Hint:

If $f(n)=3^{2n-1}+2^{n+1}$

$$f(m+1)-2f(m)=3^{2m-1}(3^2-2)$$ which is obviously divisible by $7$

So, if $7|f(m),7|f(m+1)$

If induction is not mandatory,

$$3^{2n-1}+2^{n+1}=3(3^2)^{n-1}+2^{n+1}\equiv3\cdot2^{n-1}+2^{n+1}\pmod7\equiv2^{n-1}(3+2^2)$$