Inspired by Calvin Lin - Part III

Algebra Level 5

( x + 17 ) 2 + ( 2 x + 16 ) 2 + ( 3 x + 15 ) 2 + … + ( 16 x + 2 ) 2 + ( 17 x + 1 ) 2 = 0 (x+17)^2+(2x+16)^2+(3x+15)^2+\ldots+(16x+2)^2+(17x+1)^2=0

If α \alpha and β \beta are the roots of the equation above, and suppose ∣ α + β − α β ∣ = a b | \alpha + \beta - \alpha \beta | = \frac a b for coprime positive integers a , b a,b , find the value of a + b a+b .


The answer is 108.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

( x + 17 ) 2 + ( 2 x + 16 ) 2 + ( 3 x + 15 ) 2 + . . . + ( 16 x + 2 ) 2 + ( 17 x + 1 ) 2 = 0 ∑ k = 1 17 [ k x + ( 18 − k ) ] 2 = 0 ∑ k = 1 17 [ k 2 x 2 + 2 k ( 18 − k ) x + ( 18 − k ) 2 ] 2 = 0 x 2 ∑ k = 1 17 k 2 + 2 x ( 18 ∑ k = 1 17 k − ∑ k = 1 17 k 2 ) + ∑ k = 1 17 k 2 = 0 17 ( 18 ) ( 35 ) 6 x 2 + 2 x ( 18 × 17 ( 18 ) 2 − 17 ( 18 ) ( 35 ) 6 ) + 17 ( 18 ) ( 35 ) 6 = 0 35 x 2 + 2 x ( 3 × 18 − 35 ) + 35 = 0 35 x 2 + 38 x + 35 = 0 \begin{aligned} (x+17)^2+(2x+16)^2+(3x+15)^2+...+(16x+2)^2+(17x+1)^2 & = 0 \\ \sum_{k=1}^{17} {[kx +(18-k)]^2} & = 0 \\ \sum_{k=1}^{17} {[k^2x^2 +2k(18-k)x + (18-k)^2]^2} & = 0 \\ x^2 \sum_{k=1}^{17} {k^2} + 2x \left( 18 \sum_{k=1}^{17} {k} - \sum_{k=1}^{17} {k^2} \right) + \sum_{k=1}^{17} {k^2} & = 0 \\ \frac{17(18)(35)}{6}x^2 + 2x\left( 18\times \frac{17(18)}{2} - \frac{17(18)(35)}{6} \right) +\frac{17(18)(35)}{6} & = 0 \\ 35x^2 + 2x(3\times18-35) + 35 & = 0 \\ 35x^2 + 38x + 35 & = 0 \end{aligned}

Using Vieta formulas,

α + β = − 38 35 \alpha + \beta = - \dfrac{38}{35} and α β = 1 \alpha \beta = 1

⇒ ∣ α + β − α β ∣ = ∣ − 38 35 − 1 ∣ = ∣ − 73 35 ∣ = 73 35 \Rightarrow |\alpha + \beta - \alpha \beta | = \left|- \dfrac{38}{35}-1 \right| = \left| -\dfrac{73}{35} \right| = \dfrac{73}{35}

⇒ a + b = 73 + 35 = 108 \Rightarrow a + b = 73+35 = \boxed{108}

12 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

There's an easy way to show that α β = 1 \alpha \beta = 1 , so, the majority of your working could be focused solely on finding the value of α + β \alpha + \beta .

Bonus question : Find a general formula to determine A n + B n A_n + B_n if A n A_n and B n B_n are the roots to the equation ∑ k = 1 n [ k x + ( n − k + 1 ) ] 2 = 0 \displaystyle \sum_{k=1}^n [kx + (n-k+1)]^2 = 0 .

Done. Sorry, I used a spreadsheet to do the calculations. Factorization wasn't important.

Chew-Seong Cheong - 6 years ago

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