Sum up this series

Calculus Level 3

Let T n T_n be a recurrence relation satisfying T 0 = 1 T_0 = 1 , T 1 = 3 T_1 = 3 and T n = 3 T n 1 + T n 2 T_n = 3T_{n-1} + T_{n-2} for n 2 n \ge 2 . If n = 0 T n 1 0 n = p q \displaystyle \sum_{n=0}^\infty \frac{T_n}{10^n} = \dfrac pq , where p p and q q are coprime positive integers, find p + q p + q .


The answer is 169.

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2 solutions

there is a typo

I Gede Arya Raditya Parameswara - 3 years, 5 months ago

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please indicate it

Mrigank Shekhar Pathak - 3 years, 5 months ago
Hassan Abdulla
Jan 1, 2018

Let S = n = 0 T n 10 n S=\sum _{ n=0 }^{ \infty }{ \frac { { T }_{ n } }{ { 10 }^{ n } } }

S = 1 10 0 + 3 1 0 1 + n = 2 T n 10 n S=\frac { 1 }{ { 10 }^{ 0 } } +\frac { 3 }{ 10^{ 1 } } +\sum _{ n=2 }^{ \infty }{ \frac { { T }_{ n } }{ { 10 }^{ n } } }

S = 1 10 0 + 3 1 0 1 + n = 2 ( 3 T n 1 + T n 2 10 n ) S=\frac { 1 }{ { 10 }^{ 0 } } +\frac { 3 }{ 10^{ 1 } } +\sum _{ n=2 }^{ \infty }{ \left( \frac { { 3T }_{ n-1 }+{ T }_{ n-2 } }{ { 10 }^{ n } } \right) }

S = 1 10 0 + 3 1 0 1 + 3 10 n = 2 ( T n 1 10 n 1 ) + 1 100 n = 2 ( T n 2 10 n 2 ) S=\frac { 1 }{ { 10 }^{ 0 } } +\frac { 3 }{ 10^{ 1 } } +\frac { 3 }{ 10 } \sum _{ n=2 }^{ \infty }{ \left( \frac { { T }_{ n-1 } }{ { 10 }^{ n-1 } } \right) } +\frac { 1 }{ 100 } \sum _{ n=2 }^{ \infty }{ \left( \frac { { T }_{ n-2 } }{ { 10 }^{ n-2 } } \right) }

S = 1 10 0 + 3 1 0 1 + 3 10 n = 1 ( T n 10 n ) + 1 100 n = 0 ( T n 10 n ) S=\frac { 1 }{ { 10 }^{ 0 } } +\frac { 3 }{ 10^{ 1 } } +\frac { 3 }{ 10 } \sum _{ n=1 }^{ \infty }{ \left( \frac { { T }_{ n } }{ { 10 }^{ n } } \right) } +\frac { 1 }{ 100 } \sum _{ n=0 }^{ \infty }{ \left( \frac { { T }_{ n } }{ { 10 }^{ n } } \right) }

S = 1 10 0 + 3 1 0 1 + 3 10 ( n = 0 ( T n 10 n ) 1 10 0 ) + 1 100 n = 0 ( T n 10 n ) S=\frac { 1 }{ { 10 }^{ 0 } } +\frac { 3 }{ 10^{ 1 } } +\frac { 3 }{ 10 } \left( \sum _{ n=0 }^{ \infty }{ \left( \frac { { T }_{ n } }{ { 10 }^{ n } } \right) } -\frac { 1 }{ { 10 }^{ 0 } } \right) +\frac { 1 }{ 100 } \sum _{ n=0 }^{ \infty }{ \left( \frac { { T }_{ n } }{ { 10 }^{ n } } \right) }

S = 1 + 3 10 + 3 10 ( S 1 ) + 1 100 S S=1+\frac { 3 }{ 10 } +\frac { 3 }{ 10 } \left( S-1 \right) +\frac { 1 }{ 100 } S

( 1 3 10 1 100 ) S = 1 \left( 1-\frac { 3 }{ 10 } -\frac { 1 }{ 100 } \right) S=1

S = 100 69 S=\frac { 100 }{ 69 }

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