CATAlgebra > Medium(1003)215−3(1003)2^{15} - 3(1003)215−3(997)214+3(997)2^{14}+ 3(997)214+3(1003)15+6(1003)^{15}+ 6(1003)15+6(997)15−3(997)^{15}-3(997)15−3✅ Correct Option: 1Related questions:CAT 2024 Slot 2The sum of the infinite series is 15(15−17)+(15)2((15)2−(17)2)+(15)3((15)3−(17)3)+….\frac{1}{5}\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{5}\right)^{2}\left(\left(\frac{1}{5}\right)^{2}-\left(\frac{1}{7}\right)^{2}\right)+\left(\frac{1}{5}\right)^{3}\left(\left(\frac{1}{5}\right)^{3}-\left(\frac{1}{7}\right)^{3}\right)+\ldots .51(51−71)+(51)2((51)2−(71)2)+(51)3((51)3−(71)3)+….. equal toCAT 2022 Slot 1For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n+2n2)(n + 2n^2)(n+2n2). If the nthn^{th}nth term of the progression is divisible by 999, then the smallest possible value of nnn isCAT 2024 Slot 3Consider the sequence t1=1t_1 = 1t1=1, t2=−1t_2 = -1t2=−1 and tn=(n−3n−1)tn−2t_n = \left(\frac{n-3}{n-1}\right) t_{n-2}tn=(n−1n−3)tn−2 for n≥3n \ge 3n≥3. The, the value of the sum 1t2+1t4+1t6+⋯+1t2022+1t2024\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}t21+t41+t61+⋯+t20221+t20241 is