A hideous portrayal of an old favourite

Algebra Level pending

Define the function $\displaystyle \forall z\in\mathbb{N}:f(z)=\sum_{i=1}^z\left(2^i-\frac{4a_i}{2^z}\right)^{2^i}$ , where $a\in\mathbb{R}$ .

Find $\displaystyle \sum_{i=1}^6 a_i$ if $z=6$ is a root of $f(z)$ .

Please tell me or correct any notation if it is outright wrong , but the problem itself is meant to be near unintelligible to add to the fun. =)

The answer is 2016.

1 solution

Lolly Lau
Dec 6, 2016

Since $z=6$ is a root of $f(z)$ , $f(6)=0$ .

It then follows that $\sum_{i=1}^6\left(2^i-\frac{4a_i}{2^6}\right)^{2^i}=0$ .

And if we were so brave to write it all out,

$\left(2-\frac{a_1}{16}\right)^2+\left(4-\frac{a_2}{16}\right)^4+\left(8-\frac{a_3}{16}\right)^8+\left(16-\frac{a_4}{16}\right)^{16}+\left(32-\frac{a_5}{16}\right)^{32}+\left(64-\frac{a_6}{16}\right)^{64}=0$

Since $x^{2}\geq 0$ for real $x$ , we have equality when all the terms are equal to zero.

Therefore the required sum is $32+64+128+256+512+1024=\textbf{2016}$ .

A hideous portrayal of an old favourite

Please tell me or correct any notation if it is outright wrong , but the problem itself is meant to be near unintelligible to add to the fun. =)

The answer is 2016.

1 solution

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