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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

The sequence a_n _ n ≥ 0 is defined by a_0=1, a_1=-4 and a_ n+2 =-4...

The sequence ⟨an⟩n≥0\left\langle a_n\right\rangle_{n \geq 0} is defined by a0=1,a1=−4a_0=1, a_1=-4 and an+2=−4an+1−7ana_{n+2}=-4 a_{n+1}-7 a_n, for n≥0n \geq 0. Find the number of positive integer divisors of a502−a49a51a_{50}^2-a_{49} a_{51}.

Correct answer

51

Step-by-step solution

The sequence {an},n≥0\left\{a_n\right\}, n \geq 0 and a0=1,a1=−4a_0=1, a_1=-4 and given that an+2=−4an+1−7ana_{n+2}=-4 a_{n+1}-7 a_n......(1) Now, an2−an+1⋅an−1=an2−(4an−7an−1)an−1=an2+4an⋅an−1+7an−12=−an(7an−2)+7an−12=7(an−12−an⋅an−2)=72(an−12−an−1⋅an−3)\begin{aligned} & a_n^2-a_{n+1} \cdot a_{n-1}=a_n^2-\left(4 a_n-7 a_{n-1}\right) a_{n-1} \\ & =a_n^2+4 a_n \cdot a_{n-1}+7 a_{n-1}^2 \\ & =-a_n\left(7 a_{n-2}\right)+7 a_{n-1}^2 \\ & =7\left(a_{n-1}^2-a_n \cdot a_{n-2}\right) \\ & =7^2\left(a_{n-1}^2-a_{n-1} \cdot a_{n-3}\right)\end{aligned} ................................. ∴an2−an+1an−1=7n−1(a12−a0a2)=7n−1(16−1⋅9)=7n∴a502−a49⋅a51=750\begin{array}{ll} \therefore & a_n^2-a_{n+1} a_{n-1}=7^{n-1}\left(a_1^2-a_0 a_2\right)=7^{n-1}(16-1 \cdot 9)=7^n \\ \therefore & a_{50}^2-a_{49} \cdot a_{51}=7^{50} \end{array}\therefore \quadNumberofpositiveintegerdivisorsofNumber of positive integer divisors of7^{50}=50+1=51
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