IPMAT Indore 2020Algebra > 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:Let S1={100,105,110,115,...}S_1 = \{100, 105, 110, 115, ... \}S1={100,105,110,115,...} and S2={100,95,90,85,...}S_2 = \{100, 95, 90, 85, ... \}S2={100,95,90,85,...} be two series in arithmetic progression. If aka_kak and bkb_kbk are the kkk-th terms of S1S_1S1 and S2S_2S2, respectively, then ∑k=120akbk\sum_{k=1}^{20} a_k b_k∑k=120akbk equals __________.If the sum of the first 212121 terms of the sequence: lnab,lnabb,lnab2,lnab2b,…\ln \frac{a}{b}, \ln \frac{a}{b \sqrt{b}}, \ln \frac{a}{b^{2}}, \ln \frac{a}{b^{2} \sqrt{b}}, \ldotslnba,lnbba,lnb2a,lnb2ba,… is lnambn\ln \frac{a^{m}}{b^{n}}lnbnam, then the value of m+nm+nm+n is \qquadIt 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 _______