History of Hindu Mathematics: A Source Book v2.c17.s18.8
History of Hindu Mathematics: A Source Book v2.c17.s18.8
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v2.c17.s18.8
and BC=(n-1)a. Then AD²=na². This gives the formula $a²(√n)²+a²((n-1)/2)²=a²((n+1)/2)².$
History of Hindu Mathematics: A Source Book v2.c17.s18.8
and BC=(n-1)a. Then AD²=na². This gives the formula $a²(√n)²+a²((n-1)/2)²=a²((n+1)/2)².$