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# Mathematics in Ancient and Medieval India c8.s3.43

Mathematics in Ancient and Medieval India c8.s3.43

- **Language:** English

### [c8.s3.43](https://indike.org/pramana/ins.science.bag-1979.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) $2f (n) & = 1/n - 1/(n³) + 1/(n⁵) - 1/(n⁷) + / & = n/(n² + 1)$ $Hence f (n + 1) = (n + 12)/((n + 1)² + 1)$

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Source: [https://indike.org/pramana/ins.science.bag-1979.c8.s3.43](https://indike.org/pramana/ins.science.bag-1979.c8.s3.43) · Pramāṇa
