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Is there an easy way to prove that $\operatorname{Re}(a_n) \to \infty$ where $a_n=\left(\frac{1}{2}+i\frac{\sqrt{7}}{2}\right)^n$?

Of course $|a_n| \to \infty$, but we have $$ \operatorname{Re}(a_n)=2^{-n}\sum_{0 \leq k \leq [n/2]} (-1)^k\binom{n}{k}7^k $$ where $[n/2]$ stands for the integer part of $n/2$.

However I don't see how this could be computed differently, or estimated in order to prove the claim...

Is there an easy way to prove that $|\operatorname{Re}(a_n)| \to \infty$ where $a_n=\left(\frac{1}{2}+i\frac{\sqrt{7}}{2}\right)^n$?

Of course $|a_n| \to \infty$, but we have $$ \operatorname{Re}(a_n)=2^{-n}\sum_{0 \leq k \leq [n/2]} (-1)^k\binom{n}{k}7^k $$ where $[n/2]$ stands for the integer part of $n/2$.

However I don't see how this could be computed differently, or estimated in order to prove the claim...

Is there an easy way to prove that $\operatorname{Re}(a_n) \to \infty$ where $a_n=\left|\frac{1}{2}+i\frac{\sqrt{7}}{2}\right|^n$?

Of course $|a_n| \to \infty$, but we have $$ \operatorname{Re}(a_n)=2^{-n}\sum_{0 \leq k \leq [n/2]} (-1)^k\binom{n}{k}7^k $$ where $[n/2]$ stands for the integer part of $n/2$.

However I don't see how this could be computed differently, or estimated in order to prove the claim...

Is there an easy way to prove that $\operatorname{Re}(a_n) \to \infty$ where $a_n=\left(\frac{1}{2}+i\frac{\sqrt{7}}{2}\right)^n$?

Of course $|a_n| \to \infty$, but we have $$ \operatorname{Re}(a_n)=2^{-n}\sum_{0 \leq k \leq [n/2]} (-1)^k\binom{n}{k}7^k $$ where $[n/2]$ stands for the integer part of $n/2$.

However I don't see how this could be computed differently, or estimated in order to prove the claim...

Is there an easy way to prove that $Re(a_n) \to \infty$ where $a_n=|\frac{1}{2}+i\frac{\sqrt{7}}{2}|^n$?

Of course $|a_n| \to \infty$, but we have $$Re(a_n)=2^{-n}\sum_{0 \leq k \leq [n/2]} (-1)^k\binom{n}{k}7^k$$ where $[n/2]$ stands for the integer part of $n/2$.

However I don't see how this chould be computed differently, or estimated in order to prove the claim  ...

Is there an easy way to prove that $\operatorname{Re}(a_n) \to \infty$ where $a_n=\left|\frac{1}{2}+i\frac{\sqrt{7}}{2}\right|^n$?

Of course $|a_n| \to \infty$, but we have $$ \operatorname{Re}(a_n)=2^{-n}\sum_{0 \leq k \leq [n/2]} (-1)^k\binom{n}{k}7^k $$ where $[n/2]$ stands for the integer part of $n/2$.

However I don't see how this could be computed differently, or estimated in order to prove the claim...