Working with Cubic Polynomial

Algebra Level 5

For a positive integer k ≥ 2 k\geq 2 , let a k , b k , c k a_k,b_k,c_k be the complex roots of the equation ( x − 1 k − 1 ) ( x − 1 k ) ( x − 1 k + 1 ) = 1 k \left(x-\dfrac{1}{k-1}\right)\left(x-\dfrac{1}{k}\right)\left(x-\dfrac{1}{k+1}\right)=\dfrac{1}{k} and let p k = a k ( b k + 1 ) , q k = b k ( c k + 1 ) and r k = c k ( a k + 1 ) . p_k=a_k(b_k+1), q_k = b_k(c_k+1)\text{ and } r_k = c_k(a_k+1). Given that ∑ k = 2 ∞ p k q k r k k + 1 = m n \sum_{k=2}^\infty \dfrac{p_kq_kr_k}{k+1} = \dfrac{m}{n} for some coprime positive integers m m and n n , find m + n m+n .


The answer is 43.

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Russelle Guadalupe by Russelle Guadalupe , 25, Philippines

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1 solution

By Vieta's formula, we obtain a k + b k + c k = 1 k − 1 + 1 k + 1 k + 1 = 3 k 2 − 1 k ( k 2 − 1 ) a k b k + b k c k + c k a k = 1 k ( k − 1 ) + 1 k ( k + 1 ) + 1 ( k + 1 ) ( k − 1 ) = 3 k 2 − 1 a k b k c k = 1 k ( k 2 − 1 ) + 1 k = k k 2 − 1 . \begin{aligned} a_k+b_k+c_k &= \dfrac{1}{k-1}+\dfrac{1}{k}+\dfrac{1}{k+1} = \dfrac{3k^2-1}{k(k^2-1)}\\ a_kb_k+b_kc_k+c_ka_k &= \dfrac{1}{k(k-1)}+\dfrac{1}{k(k+1)}+\dfrac{1}{(k+1)(k-1)} = \dfrac{3}{k^2-1} \\ a_kb_kc_k &= \dfrac{1}{k(k^2-1)}+\dfrac{1}{k} = \dfrac{k}{k^2-1}. \end{aligned} Then we have ( a k + 1 ) ( b k + 1 ) ( c k + 1 ) = a k b k c k + a k b k + b k c k + c k a k + a k + b k + c k + 1 = k k 2 − 1 + 3 k 2 − 1 + 3 k 2 − 1 k ( k 2 − 1 ) + 1 = k 2 + 3 k − 1 k ( k − 1 ) \begin{aligned} (a_k+1)(b_k+1)(c_k+1) &= a_kb_kc_k+a_kb_k+b_kc_k+c_ka_k+a_k+b_k+c_k+1\\ &= \dfrac{k}{k^2-1}+\dfrac{3}{k^2-1} + \dfrac{3k^2-1}{k(k^2-1)}+1\\ &= \dfrac{k^2+3k-1}{k(k-1)} \end{aligned} which gives p k q k r k = a k b k c k ( a k + 1 ) ( b k + 1 ) ( c k + 1 ) = k k 2 − 1 ⋅ k 2 + 3 k − 1 k ( k − 1 ) = k 2 + 3 k − 1 ( k + 1 ) ( k − 1 ) 2 . \begin{aligned} p_kq_kr_k &= a_kb_kc_k(a_k+1)(b_k+1)(c_k+1)\\ &= \dfrac{k}{k^2-1}\cdot \dfrac{k^2+3k-1}{k(k-1)}\\ &= \dfrac{k^2+3k-1}{(k+1)(k-1)^2}.\\ \end{aligned} Thus, ∑ k = 2 ∞ p k q k r k ( k + 1 ) = ∑ k = 2 ∞ k 2 + 3 k − 1 ( k + 1 ) 2 ( k − 1 ) 2 . \begin{aligned} \sum_{k=2}^\infty \dfrac{p_kq_kr_k}{(k+1)} = \sum_{k=2}^\infty \dfrac{k^2+3k-1}{(k+1)^2(k-1)^2}. \end{aligned} Since k 2 + 3 k − 1 ( k + 1 ) 2 ( k − 1 ) 2 = − 1 2 ( k + 1 ) − 3 4 ( k + 1 ) 2 + 1 2 ( k − 1 ) + 3 4 ( k − 1 ) 2 , \begin{aligned} \dfrac{k^2+3k-1}{(k+1)^2(k-1)^2} = -\dfrac{1}{2(k+1)}-\dfrac{3}{4(k+1)^2}+\dfrac{1}{2(k-1)}+\dfrac{3}{4(k-1)^2}, \end{aligned} we see that the given sum telescopes: ∑ k = 2 ∞ p k q k r k k + 1 = 1 2 ∑ k = 2 ∞ ( 1 k − 1 − 1 k + 1 ) + 3 4 ∑ k = 2 ∞ ( 1 ( k − 1 ) 2 − 1 ( k + 1 ) 2 ) = 1 2 ( 1 + 1 2 ) + 3 4 ( 1 + 1 4 ) = 3 4 + 15 16 = 27 16 . \begin{aligned} \sum_{k=2}^\infty \dfrac{p_kq_kr_k}{k+1} &= \dfrac{1}{2}\sum_{k=2}^\infty\left(\dfrac{1}{k-1}-\dfrac{1}{k+1}\right)+\dfrac{3}{4}\sum_{k=2}^\infty\left(\dfrac{1}{(k-1)^2}-\dfrac{1}{(k+1)^2}\right)\\ &= \dfrac{1}{2}\left(1+\dfrac{1}{2}\right)+\dfrac{3}{4}\left(1+\dfrac{1}{4}\right)\\ &= \dfrac{3}{4}+\dfrac{15}{16}=\dfrac{27}{16}. \end{aligned} Hence, we get m + n = 27 + 16 = 43 m+n=27+16=\boxed{43} .

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