Consider the sequence t 1 = 1 t_1 = 1 t 1 ​ = 1 , t 2 = − 1 t_2 = -1 t 2 ​ = − 1 , and t n = ( n − 3 n − 1 ) t n − 2 for n ≥ 3. t_n = \left(\frac{n-3}...

Question

Consider the sequence

t1=1t_1 = 1
,
t2=1t_2 = -1
, and

tn=(n3n1)tn2for n3.t_n = \left(\frac{n-3}{n-1}\right) t_{n-2} \quad \text{for } n \geq 3.
Then, the value of the sum
1t2+1t4+1t6++1t2022+1t2024\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \cdots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}
is:

Options

A.

1024144

B.

-1024144

C.

1012

D.

-1012

recursive sequencessummationproblem solvingcat 2024

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