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For a complex number $z$, let $(1+z)^{15}=a_{0}+a_{1} z+\cdots+a_{15} z^{15}$. The value of $$ \left(a_{0}-4 a_{2}+16 a_{4}+\cdots-2^{14} a_{14}\right)^{2}+\left(2 a_{1}-8 a_{3}+32 a_{5} \cdots-2^{15} a_{15}\right)^{2} $$

While I was entirely stumped by this question, I found some similar questions on here, specifically A sum involving binomial coefficients, powers and alternating signs. This did not equip me to be able to solve the above question, though, since this problem drops odd terms and even terms, alternates signs, AND has a factor of increasing powers of $2$, not to mention squares the result and then sums them.

I don't see how the sum can even be expressed as a binomial to somehow use calculus, nor do I see any algebraic manipulations; I've only managed to embarrassingly restate the problem statement as

$$ \left(\sum_{k=0}^{7} 2^{2 k}({ }^{15} C_{2 k})(-1)^k\right)^{2}+\left(\sum_{k=0}^{7} 2^{2 k+1}\left({ }^{15} C_{2 k+1}\right)(-1)^k\right)^{2} $$

Gary
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DVnyT
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  • Substitute $z=2i$ into $(1+z)^{15}$, the first $(\ldots)^2$ is the real part squared, the second $(\ldots)^2$ is the imaginary part squared. – peterwhy Jun 25 '22 at 14:14
  • To think that mentioning z was a complex number was actually hinting towards something. I would not have thought of that. Thanks! – DVnyT Jun 25 '22 at 14:17
  • Don't worry, you now know how to do this next time. The hint here is $z^2 = -4$. – peterwhy Jun 25 '22 at 14:21

2 Answers2

1

A slight variation. We consider the polynomial \begin{align*} \color{blue}{F(z)}&\color{blue}{=\left(\sum_{k=0}^7\binom{15}{2k}(-1)^k(2z)^{2k}\right)^2+\left(\sum_{k=0}^7\binom{15}{2k+1}(-1)^k(2z)^{2k+1}\right)^2}\tag{1} \end{align*} and with respect to OPs expression we need to calculate $F(1)$.

We obtain from (1) \begin{align*} \color{blue}{F(z)}&=\left(\sum_{{k=0}\atop{k\mathrm{\ even}}}^{15}\binom{15}{k}i^k(2z)^{k}\right)^2 +\left(\frac{1}{i}\sum_{{k=0}\atop{k\mathrm{\ odd}}}^{15}\binom{15}{k}i^k(2z)^{k}\right)^2\\ &=\left(\frac{(1+2iz)^{15}+(1-2iz)^{15}}{2}\right)^2+\left(\frac{1}{i}\,\frac{(1+2iz)^{15}-(1-2iz)^{15}}{2}\right)^2\\ &=\frac{1}{4}\left((1+2iz)^{15}+(1-2iz)^{15}\right)^2-\frac{1}{4}\left((1+2iz)^{15}-(1-2iz)^{15}\right)^2\\ &=\left(1+2iz\right)^{15}\left(1-2iz\right)^{15}\\ &\,\,\color{blue}{=\left(1+4z^2\right)^{15}}\\\\ \color{blue}{F(1)}&=\color{blue}{5^{15}} \end{align*} in accordance with another given answer.

Markus Scheuer
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0

Let

$$F= \left(\sum_{k=0}^{7} 2^{2 k}({ }^{15} C_{2 k})(-1)^k\right)^{2}+\left(\sum_{k=0}^{7} 2^{2 k+1}\left({ }^{15} C_{2 k+1}\right)(-1)^k\right)^{2} $$ From binomial expansion we have

$S_1=\sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k} x^{2k}=\frac{1}{2}[(1+ix)^{2n+1}+(1-ix)^{2n+1}]-(1+x^2)^{n+1/2} \cos (2n+1)t$,

$t=\tan^{-1}x.$

$S_2=\sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k+1} x^{2k+1}=\frac{-i}{2}[(1+ix)^{2n+1}-(1-ix)^{2n+1}]=(1+x^2)^{n+1/2} \sin (2n+1)t.$

$F=S_1^2+S_2^2=(1+x^2)^{2n+1},\text{where} x=2, n=7$

So we get $F=5^{15}.$

Z Ahmed
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