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Level 2

Find a+b

∑ l ( n l ) = a b cos ⁡ n π b + a n b \displaystyle \sum _{ l }^{ }{ \left( \begin{matrix} n \\ l \end{matrix} \right) } =\frac { a }{ b } \cos { \frac { n\pi }{ b } } +\frac{{ a }^{ n }}{b}

where l ≡ 0 m o d ( 3 ) \displaystyle l\equiv 0mod(3) and a,b are different primes

Note: The sum is over l such that l is 0 modulo 3 i.e. l is divisible by 3.


The answer is 5.

คลุง แจ็ค by คลุง แจ็ค , 26, United Kingdom

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1 solution

Let ω \omega be a third root of unity that is not equal to 1.

Then, by using the identity 1 + ω + ω 2 = 0 1+\omega+\omega^{2}=0 , we get

( 1 + ω ) n + ( 1 + ω 2 ) n + 2 n = 3 S (1+\omega)^{n}+(1+\omega^{2})^{n}+2^{n}=3S where S is the sum we want.

Now, by expanding ω = − 1 + 3 i 2 \omega=\dfrac{-1+\sqrt{3}i}{2} we get

S = 2 cos ⁡ n π 3 + 2 n 3 S=\dfrac { 2\cos { \dfrac { n\pi }{ 3 } } +{ 2 }^{ n } }{ 3 }

So,a=2 and b=3 and therefore, a + b = 5 a+b=\boxed{5}

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