IPMAT IndoreAlgebra > Hardπ28\frac{\pi^2}{8}8π2π216\frac{\pi^2}{16}16π2π212\frac{\pi^2}{12}12π2π236\frac{\pi^2}{36}36π2✅ Correct Option: 1Related questions:IPMAT Indore 2019Assume that all positive integers are written down consecutively from left to right as in 1234567891011...... The 6389th digit in this sequence isIPMAT Indore 2025Given that 1+122+132+142+...=π261 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}1+221+321+421+...=6π2, the value of 1+132+152+172+...1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ...1+321+521+721+... isIPMAT Indore 2021It is given that the sequence {xnx_nxn} satisfies x1=0,xn+1=xn+1+2√(1+xn)x_1 = 0, x_{n+1} = x_n + 1 + 2√(1+x_n)x1=0,xn+1=xn+1+2√(1+xn) for n=1,2,...n = 1,2,...n=1,2,... Then x31x_{31}x31 is _______