$a_1 ^n + a_2 ^n + a_3 ^n + a_4 ^n + a_5 ^n + a_6 ^n + a_7 ^n$

For positive integer $n$ , denote $S_n$ as the value of the above expression, where $a_1, a_2, a_3, a_4, a_5, a_6, a_7$ are complex numbers.

Given that $S_1 = S_2 = S_3 = S_4 = S_5 = 0$ , $S_6 = 12$ , and $\displaystyle \sum_{\text{cyclic}} a_1 a_2 a_3 a_4 a_5 a_6 = -2$

Find the value of $x$ if $S_{150} = 12x$ .

The answer is 16777216.

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You need to prove that $S_{6n} = 2^n \times 6$

Pi Han Goh
- 6 years, 2 months ago

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It can easily be observed that it's in that sequence..

Sakanksha Deo
- 6 years, 2 months ago

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You only showed that for it's true for $n = 1,2,3$ . You need a rigorous proof of that.

Pi Han Goh
- 6 years, 2 months ago

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@Pi Han Goh – Since $S_1 = P_2 = P_3 = P_4 = P_5 = 0$

So, only $S_{6n}$ and $S_{7n}$ will have non zero values.

Also every $S_{6n}$ can be evaluated by just multipliting $S_{6(n-1) }$ to - $P_6$ Or in simple sense,

$S_{6n} = S_{6(n-1) } \times ( - P_6 )= S_{6(n-1) } \times 2$

I think this should be enough

Sakanksha Deo
- 6 years, 2 months ago

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@Sakanksha Deo – Yes, you easily get interesting conclusions by observation from this since you get the following recurrence after a bit of work:

$P_i=2P_{i-6}+e_7P_{i-7}~\forall~i\geq 7~,~P_0=7$

By Newton's Identities, it can be trivially shown that,

$\forall~x\in\Bbb{Z_{\leq 5}^+}~,~P_x=0\implies e_x=0$

$\because P_x=e_x=0~\forall~x\in\Bbb{Z_{\leq 5}^+}\implies \begin{cases}P_i=0~\forall~i\not\equiv 0\pmod{6,7}\\P_i=2P_{i-6}~\forall~i\equiv 0\pmod6\\ P_{i}=e_7P_{i-7}~\forall~i\equiv 0\pmod7\end{cases}$

where $i\in\Bbb{Z^+}$ . So, you get, with a bit of recurrence solving,

$P_i=\begin{cases}2^{(i-6)/6}\times P_6=12\times 2^{(i-6)/6}~\forall~i\equiv 0\pmod6\\ e_7^{(i-7)/7}\times P_7=e_7^{(i-7)/7}\times 7e_7=7\times e_7^{i/7}~\forall~i\equiv 0\pmod7\end{cases}$

This is, in no way, a nice rigorous proof. Although, a more rigorous proof can be provided, I guess. If anyone wants, I'll try to post a more rigorous proof later here as a comment or as a separate note.

If we knew the value of $e_7$ , we could've given a simple closed form result for $P_{7n}$ values too that relied simply on $n$ .

**
Clarifications:
**

- $P_i$ and $e_i$ notation in my comment is the same as $S_i$ and $P_i$ notation in your solution respectively.
- $\Bbb{Z_{\leq 5}^+}=\{1,2,3,4,5\}$

Prasun Biswas
- 6 years, 1 month ago

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@Prasun Biswas – @Pi Han Goh , Is this rigorous enough for you? :3

Prasun Biswas
- 6 years, 1 month ago

@Prasun Biswas – Good job.....i think u sound much more technical than me.... :P

Sakanksha Deo
- 6 years, 1 month ago

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@Sakanksha Deo – Lol, it's actually quite the opposite for me. I would describe myself as more of an abstract thinker than a technical thinker.

Prasun Biswas
- 6 years, 1 month ago

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@Prasun Biswas – Well, I don't think so....by d way..i guess u r in 11th....right!!

Sakanksha Deo
- 6 years, 1 month ago

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@Sakanksha Deo – Nope, I just gave my class 12th board finals a few months ago. Waiting for the board results to be announced on 20th of this month.

Prasun Biswas
- 6 years, 1 month ago

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@Prasun Biswas – Good luck..... :)

Sakanksha Deo
- 6 years, 1 month ago

I did same.

Dev Sharma
- 5 years, 5 months ago

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Newton's sum,

Let

$\large \displaystyle\sum{a_1 a_2} = P_2$

$\large \displaystyle\sum{a_1 a_2 a_3} = P_3$

.

.

$\large a_1 a_2 a_3 a_3 a_4 a_5 a_6 a_7 = P_7$

$\large S_2 = S_1^2 - 2 P_2 \Rightarrow P_2 = 0$

Now,

$\large S_3 = S_1 S_2 - P_2 S_1 + 3 P_3 \Rightarrow P_3 = 0$

$\large S_4 = S_1 S_3 - P_2 S_2 + P_3 S_1 - 4 P4 \Rightarrow P_4 = 0$

$\large S_5 = S_1 S_4 - P_2 S_3 + P_3 S_2 - P_4 S_1 + 5 P_5 \Rightarrow P_5 = 0$

$\large S_6 = S_1 S_5 - P_2 S_4 + P_3 S_3 - P_4 S_2 + P_5 S_1 - 6 P_6 = 12 = 2^1 \times 6$

$\large S_{11} = S_1 S_{10} - P_2 S_9 + P_3 S_8 + P_4 S_7 - P_5 S_6 + P_6 S_5 - P_7 S_4 = 0$

$\large S_{12} = S_1 S_{11} - P_2 S_{10} + P_3 S_9 - P_4 S_8 + P_5 S_7 - P_6 S_6 + P_7 S_5 = 24 = 2^2 \times 6$

$\large S_{18} = S_1 S_{17} - P_2 S_{16} + P_3 S_{15} - P_4 S_{14} + P_5 S_{13} - P_6 S_{12} + P_7 S_{11} = 48 = 2^3 \times 6$

Now, notice that

$\large S_{6n} = 2^n \times 6$

Therefore,

$\large S_{150} = S_{ 6 \times 25 } = 2^{25} \times 6 = 201326592$

So,

$\large x = \frac{201326592}{12} =$

$\boxed{16777216}$