CAT 2024 Slot 3 QA Question 17

Multiple choice (+3 / −1) · Algebra · Progression & Series · Try it, then check the answer and solution below.

CAT 2024 Slot 3QAAlgebra • Progression & SeriesModerate

Consider the sequence

t_1 = 1 , t_2 = -1

and

t_n = \left(\frac{n-3}{n-1}\right) t_{n-2}

for

n \ge 3

. The, the value of the sum

\frac{1}{t_2}

\frac{1}{t_4}

\frac{1}{t_6}

+ \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}$ is

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Official Convenor Answer Key: C

Official Correct Answer: C. First, the sequence is defined as t₁ = 1, t₂ = -1, and for n ≥ 3, tₙ = ((n - 3)/(n - 1)) * tₙ₋₂. We find the sum of 1/t₂ + 1/t₄ + 1/t₆ + ... up to 1/t₂₀₂₄. Hmm, okay. So, the sequence is defined recursively, and each term depends on the term two places before it. That makes me think it's some kind of alternating or periodic sequence. Maybe I can find a pattern or a general formula for tₙ. We have the first few terms to see if I can spot a pattern. We know t₁ = 1 and t₂ = -1. For n = 3: t₃ = ((3 - 3)/(3 - 1)) * t₁ = (0/2) * 1 = 0. So, t₃ = 0. For n = 4: t₄ = ((4 - 3)/(4 - 1)) * t₂ = (1/3) * (-1) = -1/3. For n = 5: t₅ = ((5 - 3)/(5 - 1)) * t₃ = (2/4) * 0 = 0. For n = 6: t₆ = ((6 - 3)/(6 - 1)) * t₄ = (3/5) * (-1/3) = (-1)/5. Wait, so t₆ is -1/5. Let me keep going a bit more. n = 7: t₇ = ((7 - 3)/(7 - 1)) * t₅ = (4/6) * 0 = 0. n = 8: t₈ = ((8 - 3)/(8 - 1)) * t₆ = (5/7) * (-1/5) = (-1)/7. Hmm, interesting. It seems like for even n, the terms are non-zero, and for odd n, starting from t₃, they are zero. So, t₁ = 1, t₂ = -1, t₃ = 0, t₄ = -1/3, t₅ = 0, t₆ = -1/5, t₇ = 0, t₈ = -1/7, and so on. So, the non-zero terms are at even indices: t₂, t₄, t₆, etc., each equal to -1/(n-1) where n is the index. Wait, let's check: t₂ = -1 = -1/(2 - 1) = -1/1 = -1. Correct. t₄ = -1/3 = -1/(4 - 1). Correct. t₆ = -1/5 = -1/(6 - 1). Correct. t₈ = -1/7 = -1/(8 - 1). Correct. So, in general, for even n, tₙ = -1/(n - 1). That seems to be the pattern. So, if I can express tₙ for even n as -1/(n - 1), then 1/tₙ would be - (n - 1). Therefore, the sum we need is the sum of 1/t₂ + 1/t₄ + ... + 1/t₂₀₂₄, which translates to: Sum = (- (2 - 1)) + (- (4 - 1)) + (- (6 - 1)) + ... + (- (2024 - 1)). Simplifying each term: - (2 - 1) = -1 - (4 - 1) = -3 - (6 - 1) = -5 ... - (2024 - 1) = -2023 So, the sum becomes: Sum = (-1) + (-3) + (-5) + ... + (-2023) Which is the same as: Sum = - (1 + 3 + 5 + ... + 2023) Now, I need to compute the sum of the first few odd numbers up to 2023 and then multiply by -1. Wait, how many terms are there in this sequence? . The sequence is 1, 3, 5, ..., 2023. This is an arithmetic sequence with first term a = 1, common difference d = 2. The nth term of an arithmetic sequence is given by aₙ = a + (n - 1)d. So, 2023 = 1 + (n - 1)*2 2023 - 1 = (n - 1)*2 2022 = 2(n - 1) n - 1 = 1011 n = 1012 So, there are 1012 terms in this sequence. The sum of an arithmetic series is given by S = n/2 * (a + l), where l is the last term. So, S = 1012/2 * (1 + 2023) = 506 * 2024 Compute 506 * 2024. First, note that 506 * 2000 = 1,012,000 Then, 506 * 24 = 12,144 So, total sum S = 1,012,000 + 12,144 = 1,024,144 Therefore, the sum we're looking for is -1,024,144. Looking at the options, option C is -1024144, which matches. 1. Found the first few terms and noticed a pattern where even terms are -1/(n-1) and odd terms beyond t₁ are zero. 2. Expressed 1/tₙ for even n as - (n - 1). 3. Sum becomes the sum of -1, -3, -5, ..., -2023, which is - (sum of 1 + 3 + 5 + ... + 2023). Correct. 4. Calculated the number of terms: 1012. Correct, since 2023 is the 1012th odd number. 5. Sum of the series: 1012/2 * (1 + 2

Answer and solution

Answer: C) -1024144

First, the sequence is defined as t₁ = 1, t₂ = -1, and for n ≥ 3, tₙ = ((n - 3)/(n - 1)) * tₙ₋₂. We find the sum of 1/t₂ + 1/t₄ + 1/t₆ + ... up to 1/t₂₀₂₄.

Hmm, okay. So, the sequence is defined recursively, and each term depends on the term two places before it. That makes me think it's some kind of alternating or periodic sequence. Maybe I can find a pattern or a general formula for tₙ.

We have the first few terms to see if I can spot a pattern.

We know t₁ = 1 and t₂ = -1.

For n = 3: t₃ = ((3 - 3)/(3 - 1)) t₁ = (0/2) 1 = 0. So, t₃ = 0.

For n = 4: t₄ = ((4 - 3)/(4 - 1)) t₂ = (1/3) (-1) = -1/3.

For n = 5: t₅ = ((5 - 3)/(5 - 1)) t₃ = (2/4) 0 = 0.

For n = 6: t₆ = ((6 - 3)/(6 - 1)) t₄ = (3/5) (-1/3) = (-1)/5.

Wait, so t₆ is -1/5. Let me keep going a bit more.

n = 7: t₇ = ((7 - 3)/(7 - 1)) t₅ = (4/6) 0 = 0.

n = 8: t₈ = ((8 - 3)/(8 - 1)) t₆ = (5/7) (-1/5) = (-1)/7.

Hmm, interesting. It seems like for even n, the terms are non-zero, and for odd n, starting from t₃, they are zero. So, t₁ = 1, t₂ = -1, t₃ = 0, t₄ = -1/3, t₅ = 0, t₆ = -1/5, t₇ = 0, t₈ = -1/7, and so on.

So, the non-zero terms are at even indices: t₂, t₄, t₆, etc., each equal to -1/(n-1) where n is the index. Wait, let's check:

t₂ = -1 = -1/(2 - 1) = -1/1 = -1. Correct.

t₄ = -1/3 = -1/(4 - 1). Correct.

t₆ = -1/5 = -1/(6 - 1). Correct.

t₈ = -1/7 = -1/(8 - 1). Correct.

So, in general, for even n, tₙ = -1/(n - 1). That seems to be the pattern.

So, if I can express tₙ for even n as -1/(n - 1), then 1/tₙ would be - (n - 1).

Therefore, the sum we need is the sum of 1/t₂ + 1/t₄ + ... + 1/t₂₀₂₄, which translates to:

Sum = (- (2 - 1)) + (- (4 - 1)) + (- (6 - 1)) + ... + (- (2024 - 1)).

Simplifying each term:

(2 - 1) = -1

(4 - 1) = -3

(6 - 1) = -5

...

(2024 - 1) = -2023

So, the sum becomes:

Sum = (-1) + (-3) + (-5) + ... + (-2023)

Which is the same as:

Sum = - (1 + 3 + 5 + ... + 2023)

Now, I need to compute the sum of the first few odd numbers up to 2023 and then multiply by -1.

Wait, how many terms are there in this sequence? .

The sequence is 1, 3, 5, ..., 2023. This is an arithmetic sequence with first term a = 1, common difference d = 2.

The nth term of an arithmetic sequence is given by aₙ = a + (n - 1)d.

So, 2023 = 1 + (n - 1)*2

2023 - 1 = (n - 1)*2

2022 = 2(n - 1)

n - 1 = 1011

n = 1012

So, there are 1012 terms in this sequence.

The sum of an arithmetic series is given by S = n/2 * (a + l), where l is the last term.

So, S = 1012/2 (1 + 2023) = 506 2024

Compute 506 * 2024.

First, note that 506 * 2000 = 1,012,000

Then, 506 * 24 = 12,144

So, total sum S = 1,012,000 + 12,144 = 1,024,144

Therefore, the sum we're looking for is -1,024,144.

Looking at the options, option C is -1024144, which matches.

Found the first few terms and noticed a pattern where even terms are -1/(n-1) and odd terms beyond t₁ are zero.

Expressed 1/tₙ for even n as - (n - 1).

Sum becomes the sum of -1, -3, -5, ..., -2023, which is - (sum of 1 + 3 + 5 + ... + 2023). Correct.

Calculated the number of terms:

1012

Correct, since 2023 is the 1012th odd number.

Sum of the series: 1012/2 * (1 + 2

CAT 2024 Slot 3 QA Question 17

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