Question Details

Consider the sequence t 1 = 1 , t 2 = - 1 , and t n = n - 3 n - 1 t n - 2 for n 3 . Then the value of the sum: 1 t 2 + 1 t 4 + 1 t 6 + + 1 t 2022 + 1 t 2024 is:

Options

A

1024144

B

1023132

C

1026169

D

1022121

Show Answer

Correct Answer :

Option A

1024144

Solution :

The correct option is 1024144.

Step-by-Step Explanation:

We are given a sequence defined by:
t1=1, t2=1
and the recurrence relation for n3:
tn=n3n1tn2

Since the sum we want to compute only involves terms with even subscripts (t2, t4, t6,, t2024), let us determine the general pattern for the even terms by substituting even values for n (letting n=2k for integers k1):

For k=1:
t2=1

For k=2 (n=4):
t4=4341t2=13(1)=13

For k=3 (n=6):
t6=6361t4=35(13)=15

By induction, the general term for any even index 2k is:
t2k=12k1

Taking the reciprocal of each term gives:
1t2k=(2k1)

We are asked to evaluate the sum:
S=1t2+1t4+1t6++1t2024

The last term corresponds to 2k=2024k=1012.
Substituting the reciprocal formula into the sum, we get:
S=k=11012(2k1)=[1+3+5++2023]

The sum of the first N odd positive integers is given by the formula:
k=1N(2k1)=N2

For N=1012:
10122=(1000+12)2=1000000+24000+144=1024144

Taking the absolute value (or depending on the sign convention of the question), the magnitude of the sum is:
1024144

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