Find: sum_(n=0)^infty (1/(1+3*(-1)^n))^n

Alvin Parks

Alvin Parks

Answered question

2022-11-07

Find: n = 0 ( 1 1 + 3 ( 1 ) n ) n

Answer & Explanation

lavarcar2d2

lavarcar2d2

Beginner2022-11-08Added 18 answers

For even n, the sum is
n = 0 1 / 4 2 n = n = 0 1 / 16 n = 1 / ( 1 1 / 16 ) = 16 / 15
For odd n, the sum is
n = 0 1 / ( 2 ) 2 n + 1 = n = 0 1 / 2 2 n + 1 = ( 1 / 2 ) n = 0 1 / 2 2 n = ( 1 / 2 ) n = 0 1 / 4 n = ( 1 / 2 ) / ( 1 1 / 4 ) = ( 1 / 2 ) ( 4 / 3 ) = 2 / 3
Adding, the overall sum is 16 / 15 2 / 3 = ( 16 2 5 ) / 15 = 6 / 15 = 2 / 5

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