This one is for Mi!

Algebra Level 5

If { a 1 , a 2 , a 3 , … , a n } \{a_1,a_2,a_3,\ldots,a_n\} is a set of real numbers, placed so that a 1 < a 2 < a 3 < ⋯ < a n , a_1 < a_2 < a_3 < \cdots < a_n, its complex power sum is defined as a 1 i + a 2 i 2 + a 3 i 3 + ⋯ + a n i n , a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n, where i 2 = − 1. i^2 = - 1. Let S n S_n be the sum of the complex power sums of all nonempty subsets of { 1 , 2 , … , n } . \{1,2,\ldots,n\}. Given that S 8 = − 176 − 64 i S_8 = - 176 - 64i and S 9 = p + q i , S_9 = p + qi, where p p , q q ∈ \in Z \mathbb{Z} , find ∣ p ∣ + ∣ q ∣ |p| + |q| .

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I am posting this ques. because I have lost the bet I had with shubhendra.


The answer is 368.

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Parth Lohomi by Parth Lohomi , 20, India

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1 solution

Gian Sanjaya
May 6, 2015

S n + 1 S_{n+1} has been related to S n S_n . from the definition, S n + 1 − S n S_{n+1}-S_n is equal to the sum of the complex power sums of all subsets of { 1 , 2 , 3 , . . . , n + 1 } \{1, 2, 3, ..., n+1\} that contain n + 1 n+1 . Because n + 1 n+1 is the biggest number... S n + 1 − S n = n i + ( 1 i + n i 2 ) + ( 2 i + n i 2 ) + … S_{n+1}-S_n=ni+(1i+ni^2)+(2i+ni^2)+\ldots and so on, also because of the definition given, we get: S n + 1 − S n = S n + ( n + 1 ) i + ( n + 1 ) i ( n i ) + ( n + 1 ) i ( n C 2 ) ( i 2 ) + ⋯ + ( n + 1 ) i ( n C n ) ( i n ) S_{n+1}-S_n=S_n+(n+1)i+(n+1)i(ni)+(n+1)i(nC2)(i^2)+\cdots+(n+1)i(nCn)(i^n)
Then, S n + 1 − 2 S n = n i ( 1 + i ) n ) S_{n+1}-2S_n=ni(1+i)^n) and we get the recursive relation (isn't this recursive) S n + 1 = 2 S n + n i ( 1 + i ) n S_{n+1} = 2S_n+ni(1+i)^n Put n = 8 n=8 and we get: S 9 = 2 S 8 + 9 i ( 1 + i ) 8 S_9 = 2S_8+9i(1+i)^8 S 9 = 2 ( − 176 − 64 i ) + ( 9 i ) 16 S_9 = 2(-176-64i)+(9i)16 S 9 = − 352 − 128 i + 144 i S_9 = -352-128i+144i S 9 = − 352 + 16 i S_9 = -352+16i So, p = − 352 , q = 16 p = -352, q = 16 , and then ∣ p ∣ + ∣ q ∣ = 368 |p|+|q|=\boxed{368} . Sorry for bad writing. Anyone can post a new, better one?

Mod: LaTeX {\LaTeX} 'ed
1 Helpful 0 Interesting 0 Brilliant 0 Confused

Ah, I didn't realize that it was a binomial expansion of ( i + 1 ) 8 (i+1)^8 . Nice solution!

Daniel Liu - 5 years, 11 months ago

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