I Don't Think the a Should be There

Calculus Level 5

∑ x = 1 ∞ log ⁡ 33 ( ( x + a − 1 ) ( x + a + 1 ) ( x + a − 3 ) ( x + a + 3 ) ) = 1 \displaystyle \sum_{x=1}^{\infty} \displaystyle \log_{33}\left(\frac{(x+a-1)(x+a+1)}{(x+a-3)(x+a+3)}\right)=1

If the sum of all possible real values of constant a a such that the infinite sum above is fulfilled can be represented as p q \frac{p}{q} for coprime positive integers p , q p,q , find p + q p+q .

Try the harder (way harder) version here .

Try the easier version here .


The answer is 7.

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Trevor Arashiro by Trevor Arashiro , 22, USA

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1 solution

Chew-Seong Cheong
May 10, 2015

∑ x = 1 ∞ log ⁡ 33 ( ( x + a − 1 ) ( x + a + 1 ) ( x + a − 3 ) ( x + a + 3 ) ) = log ⁡ 33 ( ( a ) ( a + 2 ) ( a − 2 ) ( a + 4 ) × ( a + 1 ) ( a + 3 ) ( a − 1 ) ( a + 5 ) × ( a + 2 ) ( a + 4 ) ( a ) ( a + 6 ) × ( a + 3 ) ( a + 5 ) ( a + 1 ) ( a + 7 ) × ( a + 4 ) ( a + 6 ) ( a + 2 ) ( a + 8 ) × ( a + 5 ) ( a + 7 ) ( a + 3 ) ( a + 9 ) × . . . ) = log ⁡ 33 ( ( a + 2 ) ( a + 3 ) ( a − 2 ) ( a − 1 ) ) = 1 ⇒ ( ( a + 2 ) ( a + 3 ) ( a − 2 ) ( a − 1 ) ) = 33 a 2 + 5 a + 6 = 33 ( a 2 − 3 a + 2 ) a 2 + 5 a + 6 = 33 a 2 − 99 a + 66 32 a 2 − 104 a + 60 = 0 8 a 2 − 26 a + 15 = 0 ( 2 a − 5 ) ( 4 a − 3 ) = 0 ⇒ { a = 5 2 a = 3 4 unacceptable, ln ⁡ ( a − 1 ) and ln ⁡ ( a − 2 ) are undefined \displaystyle \sum_{x=1}^\infty {\log_{33} \left(\frac{(x+a-1)(x+a+1)}{(x+a-3)(x+a+3)} \right)} \\ = \log_{33} {\left(\dfrac{\color{#D61F06}{(a)}(a+2)}{(a-2)\color{#3D99F6}{(a+4)}} \times \dfrac{\color{#D61F06}{(a+1)}(a+3)}{(a-1)\color{#3D99F6}{(a+5)}} \times \dfrac{\color{#D61F06}{(a+2)}\color{#3D99F6}{(a+4)}}{\color{#D61F06}{(a)}\color{#3D99F6}{(a+6)}} \\ \quad \quad \quad \quad \quad \times \dfrac{\color{#D61F06}{(a+3)}\color{#3D99F6}{(a+5)}}{\color{#D61F06}{(a+1)}\color{#3D99F6}{(a+7)}} \times \dfrac{\color{#D61F06}{(a+4)}\color{#3D99F6}{(a+6)}}{\color{#D61F06}{(a+2)}\color{#3D99F6}{(a+8)}} \times \dfrac{\color{#D61F06}{(a+5)}\color{#3D99F6}{(a+7)}}{\color{#D61F06}{(a+3)}\color{#3D99F6}{(a+9)}} \times ... \right)} \\ = \log_{33} {\left( \dfrac{(a+2)(a+3)}{(a-2)(a-1)} \right)} = 1 \\ \Rightarrow \displaystyle \left(\frac{(a+2)(a+3)}{(a-2)(a-1)} \right) = 33 \\ \quad a^2 + 5a + 6 = 33(a^2-3a+2) \\ \quad a^2 + 5a + 6 = 33a^2-99a+66 \\ \quad 32a^2-104a+60 = 0 \\ \quad 8a^2-26a+15 = 0 \\ \quad (2a-5)(4a-3) = 0 \\ \Rightarrow \begin{cases} a = \frac{5}{2} \\ a = \frac{3}{4} \quad \text{unacceptable, } \ln{(a-1)} \text{ and } \ln{(a-2)} \text{ are undefined} \end{cases}

Therefore, the sum of all possible real values of a a is 5 2 ⇒ p + q = 5 + 2 = 7 \frac {5}{2} \quad \Rightarrow p+q = 5 + 2 = \boxed{7}

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It's not quite enough to say that ln ⁡ ( a − 1 ) \ln(a-1) and ln ⁡ ( a − 2 ) \ln(a-2) are undefined. You need that one of the terms of the sum is undefined, since the log properties you use assume that each log is separately defined. (E.g. ln ⁡ ( − 2 ⋅ − 3 ) \ln(-2 \cdot -3) is defined even though ln ⁡ ( − 2 ) \ln(-2) and ln ⁡ ( − 3 ) \ln(-3) are undefined.)

Patrick Corn - 6 years, 1 month ago

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