Mathematics in Ancient and Medieval India c8.s3.43

c8.s3.43

*Case i* ) Putting p = 2 in (8), we obtain from (6) and (8), $a₁ = 1, a₃ = - 1, a₂ = a₄ = 0.$ Now from (6) $2f (n) & = 1/n - 1/(n³) + 1/(n⁵) - 1/(n⁷) + / & = n/(n² + 1)$ $Hence f (n + 1) = (n + 12)/((n + 1)² + 1)$