Prove that $4^n \sum_{k=0}^{n} \binom nk +14n-1 $ is divisible by $7$ for every $n \geq 1$.
Basic Step: For $n=1$, $21$ is divisible by $7$.($21 \mod 7 = 0$)
Induction Hypothesis: Suppose that for an arbitrarily chosen $n \geq 1$ the number $4^n \sum_{k=0}^{n} \binom nk +14n-1 $ is divisible by $7$.
Induction Step: We will show that $4^{n+1}\sum_{k=0}^{n+1} \binom {n+1}k +14(n+1)-1 $ is divisible by 7.
\begin{align*}
4^{n+1}\sum_{k=0}^{n+1} \binom {n+1}k +14(n+1)-1 &= 4 \times 4^n \sum_{k=0}^{n} \binom{n}{k}+\binom{n+1}{n} +14n+14-1 \\
&=4 \times 4^n \sum_{k=0}^{n} \binom{n}{k}+(n+1) +14n+14-1 \\
&= 4 \times 4^n \sum_{k=0}^{n} \binom{n}{k}+14n-1+(n+1)+14
\end{align*}
So from the hypothesis we made the first term is divisible by 7. How can i show that the rest is divisible by 7? Also I am not sure if my approach is correct!
Doing arithmetic modulo $\;7\;$ :
$$4^n\sum_{k=0}^n\binom nk+14n-1=4^n\cdot2^n+14n-1=4^{n-1}2^{n-1}\cdot8+14(n-1)+14-1=$$
$$=\overbrace{\color{red}{4^{n-1}2^{n-1}+14(n-1)-1}}^{=0\;,\;\text{by Ind. Hypot.}}+\underbrace{7\left(4^{n-1}2^{n-1}+2\right)}_{=0\;\text{trivially}}$$
and we're done.
Observe that "$=0$" above means the expression equals zero modulo $\;7\;$ , i.e.: it is divisible by seven.