Mathematics in Ancient and Medieval India c8.s3.46

c8.s3.46

*Case ii* ) Putting p = 3, we obtain as before from ( 6 ) and ( 8 ) $a₁ = 1, a₃ = - 1, a₅ = 5, a₂ = a₄ = a₆ = 0$ From (6), $2f (n) & = 1/n - 1/(n³) + 5/(n⁵) - (5²)/(n⁷) + / & = (n² + 4)/(n(n² + 5))$ $Hence f (n + 1) = ((n + 1)²4 + 1)/((n + 12) [ (n + 1)² + 4 + 1 ])$