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CAT 2024 Slot 3 — QA Question 22

Miscellaneous ProgressionsEasy

Consider the sequence t1 = 1, t2 = -1 and tn = (n-3n-1)tn-2 for n ≥ 3. Then, the value of the sum 1t2 + 1t4 + 1t6 + ... + 1t2022 + 1t2024

Answer & solution

  • -1024144

  • B

    -1022121

  • C

    -1026169

  • D

    -1023132

Solution

Hard

The recurrence tn=n−3n−1 tn−2t_n=\dfrac{n-3}{n-1}\,t_{n-2} links every even-indexed term to the previous even one. Telescope the product to get a closed form for t2kt_{2k}, then 1/t2k1/t_{2k} becomes a clean quadratic in kk, and the sum is a standard sum of squares-type series.

1

Closed form for even terms. With t2=−1t_2=-1 and tn=n−3n−1tn−2t_n=\dfrac{n-3}{n-1}t_{n-2}, set n=2kn=2k and telescope the product from t2t_2.

t2k=t2∏j=2k2j−32j−1⇒ t2k=−1⋅13⋅35⋯2k−32k−1(write terms)⇒ t2k=−12k−1(telescoping cancellation)\begin{aligned} &t_{2k}=t_2\prod_{j=2}^{k}\frac{2j-3}{2j-1}\\ &\Rightarrow\ t_{2k}=-1\cdot\frac{1}{3}\cdot\frac{3}{5}\cdots\frac{2k-3}{2k-1}\quad\text{(write terms)}\\ &\Rightarrow\ t_{2k}=-\frac{1}{2k-1}\quad\text{(telescoping cancellation)} \end{aligned}
2

Reciprocal term. Hence each summand is linear in kk.

1t2k=−(2k−1)\begin{aligned} &\frac{1}{t_{2k}}=-(2k-1) \end{aligned}
3

Set the range. The sum runs over t2,t4,…,t2024t_2,t_4,\ldots,t_{2024}, i.e. k=1,2,…,1012k=1,2,\ldots,1012.

S=∑k=110121t2k=−∑k=11012(2k−1)\begin{aligned} &S=\sum_{k=1}^{1012}\frac{1}{t_{2k}}=-\sum_{k=1}^{1012}(2k-1) \end{aligned}
4

Sum the odd numbers. The sum of the first NN odd numbers is N2N^2, with N=1012N=1012.

∑k=11012(2k−1)=10122⇒ 10122=1024144(1012×1012)⇒ S=−1024144\begin{aligned} &\sum_{k=1}^{1012}(2k-1)=1012^2\\ &\Rightarrow\ 1012^2=1024144\quad\text{(}1012\times1012\text{)}\\ &\Rightarrow\ S=-1024144 \end{aligned}
−1024144-1024144
CAT 2024 Slot 3 QA Q22: 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 — Solution | TheCATExam