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Quantitative Aptitude Question on Sequence and Series

Consider the sequencet1=1,t2=−1andtn=(n−3n−1)tn−2forn≥3.t_1 = 1, t_2 = -1 and t_n = \left( \frac{n-3}{n-1} \right) t_{n-2} for n \ge 3.Then, the value of the sum 1t2\frac{1}{t_2} + 1t4\frac{1}{t_4} + 1t6\frac{1}{t_6} + ……. + 1t2022\frac{1}{t_{2022}} + 1t2024\frac{1}{t_{2024}}, is

A

-1024144

B

-1023132

C

-1026169

D

-1022121

Answer

-1024144

Explanation

Solution

We are given the recurrence relation tn=n−3n−1tn−2t_n = \frac{n-3}{n-1} t_{n-2} for n≥3n \ge 3, along with the initial terms t1=1t_1 = 1 and t2=−1t_2 = -1.
We need to calculate the sum:
S=1t2+1t4+1t6+⋯+1t2022+1t2024S = \frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}

We first observe the pattern in the terms generated by the recurrence relation. Using the recurrence, we can calculate the first few terms:
t_3$$= \frac{3-3}{3-1} t_1 = 0

t4=4−34−1t2=13×(−1)=−13t_4 = \frac{4-3}{4-1} t_2 = \frac{1}{3} \times (-1) = -\frac{1}{3}

t5=5−35−1t3=24×0=0t_5 = \frac{5-3}{5-1} t_3 = \frac{2}{4} \times 0 = 0

t_6 = \frac{6-3}{6-1} t_4 = \frac{3}{5}$$\times \left(-\frac{1}{3}\right) = -\frac{1}{5}

From this, we notice that tnt_n for even nn follows the pattern:
t2=−1,t4=−13,t6=−15,…t_2 = -1, \quad t_4 = -\frac{1}{3}, \quad t_6 = -\frac{1}{5}, \dots
Thus, the values of tnt_n for even nn are the negative reciprocals of the odd numbers starting from 1, i.e., tn=−1n−1t_n = -\frac{1}{n-1}.

Now, the sum is:
S=∑k=110121t2k=∑k=11012−(2k−1)=−∑k=11012(2k−1)S = \sum_{k=1}^{1012} \frac{1}{t_{2k}} = \sum_{k=1}^{1012} -(2k-1) = -\sum_{k=1}^{1012} (2k-1)

The sum of the first 1012 odd numbers is 101221012^2, so:
S=−10122=−1024144S = -1012^2 = -1024144
Thus, the correct answer is Option (1).