IPMAT IndoreAlgebra > Hard-343026-38✅ Correct Option: 1Related questions:IPMAT Indore 2020If 112+122+132+…\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots121+221+321+… up to ∞=π26\infty = \frac{\pi^2}{6}∞=6π2, then the value of 112+132+152+…\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots121+321+521+… up to ∞\infty∞ 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 _______IPMAT Indore 2022A new sequence is obtained from the sequence of positive integers (1,2,3,…)(1,2,3, \ldots)(1,2,3,…) by deleting all the perfect squares. Then the 2022nd 2022^{\text {nd }}2022nd term of the new sequence is ________.