A = ( a a + b b ) − n F ( 1 2 n , 1 2 n + 1 2 , 1 , 4 a a b b ( a a + b b ) 2 ) A ′ = n ( a a + b b ) − n − 1 a b F ( 1 2 n + 1 2 , 1 2 n + 1 , 2 , 4 a a b b ( a a + b b ) 2 ) A ′ ′ = n ( n + 1 ) 1.2 ( a a + b b ) − n − 2 a a b b F ( 1 2 n + 1 , 1 2 n + 3 2 , 3 , 4 a a b b ( a a + b b ) 2 ) A ′ ′ ′ = n ( n + 1 ) ( n + 2 ) 1.2.3 ( a a + b b ) − n − 3 a 3 b 3 F ( 1 2 n + 3 2 , 1 2 n + 2 , 4 , 4 a a b b ( a a + b b ) 2 ) {\displaystyle {\begin{aligned}&A=(aa+bb)^{-n}F({\tfrac {1}{2}}n,{\tfrac {1}{2}}n+{\tfrac {1}{2}},1,{\tfrac {4aabb}{(aa+bb)^{2}}})\\&A^{\prime }=n(aa+bb)^{-n-1}abF({\tfrac {1}{2}}n+{\tfrac {1}{2}},{\tfrac {1}{2}}n+1,2,{\tfrac {4aabb}{(aa+bb)^{2}}})\\&A^{\prime \prime }={\tfrac {n(n+1)}{1.2}}(aa+bb)^{-n-2}aabbF({\tfrac {1}{2}}n+1,{\tfrac {1}{2}}n+{\tfrac {3}{2}},3,{\tfrac {4aabb}{(aa+bb)^{2}}})\\&A^{\prime \prime \prime }={\tfrac {n(n+1)(n+2)}{1.2.3}}(aa+bb)^{-n-3}a^{3}b^{3}F({\tfrac {1}{2}}n+{\tfrac {3}{2}},{\tfrac {1}{2}}n+2,4,{\tfrac {4aabb}{(aa+bb)^{2}}})\\\end{aligned}}}
qui valores facile deducuntur ex
Tertio fit
A = ( a + b ) − 2 n F ( n , 1 2 , 1 , 4 a b ( a + b ) 2 ) A ′ = n ( a + b ) − 2 n − 2 a b F ( n + 1 , 3 2 , 3 , 4 a b ( a + b ) 2 ) A ′ ′ = n ( n + 1 ) 1.2 ( a + b ) − 2 n − 4 a a b b F ( n + 2 , 5 2 , 5 , 4 a b ( a + b ) 2 ) A ′ ′ ′ = n ( n + 1 ) ( n + 2 ) 1.2.3 ( a + b ) − 2 n − 6 a 3 b 3 F ( n + 3 , 7 2 , 7 , 4 a b ( a + b ) 2 ) {\displaystyle {\begin{alignedat}{2}&A&&=(a+b)^{-2n}F(n,{\tfrac {1}{2}},1,{\tfrac {4ab}{(a+b)^{2}}})\\&A^{\prime }&&=n(a+b)^{-2n-2}abF(n+1,{\tfrac {3}{2}},3,{\tfrac {4ab}{(a+b)^{2}}})\\&A^{\prime \prime }&&={\tfrac {n(n+1)}{1.2}}(a+b)^{-2n-4}aabbF(n+2,{\tfrac {5}{2}},5,{\tfrac {4ab}{(a+b)^{2}}})\\&A^{\prime \prime \prime }&&={\tfrac {n(n+1)(n+2)}{1.2.3}}(a+b)^{-2n-6}a^{3}b^{3}F(n+3,{\tfrac {7}{2}},7,{\tfrac {4ab}{(a+b)^{2}}})\\\end{alignedat}}}
Denique fit quarto
A = ( a − b ) − 2 n F ( n , 1 2 , 1 , − 4 a b ( a − b ) 2 ) A ′ = n ( a − b ) − 2 n − 2 a b F ( n + 1 , 3 2 , 3 , − 4 a b ( a − b ) 2 ) A ′ ′ = n ( n + 1 ) 1.2 ( a − b ) − 2 n − 4 a a b b F ( n + 2 , 5 2 , 5 , − 4 a b ( a − b ) 2 ) A ′ ′ ′ = n ( n + 1 ) ( n + 2 ) 1.2.3 ( a − b ) 2 n − 6 a 3 b 3 F ( n + 3 , 7 2 , 7 , − 4 a b ( a − b ) 2 ) {\displaystyle {\begin{alignedat}{2}&A&&=(a-b)^{-2n}F(n,{\tfrac {1}{2}},1,-{\tfrac {4ab}{(a-b)^{2}}})\\&A^{\prime }&&=n(a-b)^{-2n-2}abF(n+1,{\tfrac {3}{2}},3,-{\tfrac {4ab}{(a-b)^{2}}})\\&A^{\prime \prime }&&={\tfrac {n(n+1)}{1.2}}(a-b)^{-2n-4}aabbF(n+2,{\tfrac {5}{2}},5,-{\tfrac {4ab}{(a-b)^{2}}})\\&A^{\prime \prime \prime }&&={\tfrac {n(n+1)(n+2)}{1.2.3}}(a-b)^{2n-6}a^{3}b^{3}F(n+3,{\tfrac {7}{2}},7,-{\tfrac {4ab}{(a-b)^{2}}})\end{alignedat}}}
Valores illi atque hi facile eruuntur ex
