hat
Egzanp ∏ k = 1 n exp ⁡ ( k 2 ) = exp ⁡ ( ∑ k = 1 n k 2 ) = exp ⁡ ( n ( n + 1 ) ( 2 n + 1 ) 6 ) {\displaystyle \prod _{k=1}^{n}\exp(k^{2})=\exp \left(\sum _{k=1}^{n}k^{2}\right)=\exp \left({\frac {n(n+1)(2n+1)}{6}}\right)} Nou te ka ekri nan yon fason mwens kondanse : exp ⁡ ( 1 2 ) ⋅ exp ⁡ ( 2 2 ) ⋅ exp ⁡ ( 3 2 ) ⋅ … ⋅ exp ⁡ ( n 2 ) = exp ⁡ ( n ( n + 1 ) ( 2 n + 1 ) 6 ) {\displaystyle \exp(1^{2})\cdot \exp(2^{2})\cdot \exp(3^{2})\cdot \ldots \cdot \exp(n^{2})=\exp \left({\frac {n(n+1)(2n+1)}{6}}\right)}
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Exemple ∏ k = 1 n exp ⁡ ( k 2 ) = exp ⁡ ( ∑ k = 1 n k 2 ) = exp ⁡ ( n ( n + 1 ) ( 2 n + 1 ) 6 ) {\displaystyle \prod _{k=1}^{n}\exp(k^{2})=\exp \left(\sum _{k=1}^{n}k^{2}\right)=\exp \left({\frac {n(n+1)(2n+1)}{6}}\right)} On aurait pu écrire de manière moins condensée : exp ⁡ ( 1 2 ) ⋅ exp ⁡ ( 2 2 ) ⋅ exp ⁡ ( 3 2 ) ⋅ … ⋅ exp ⁡ ( n 2 ) = exp ⁡ ( n ( n + 1 ) ( 2 n + 1 ) 6 ) {\displaystyle \exp(1^{2})\cdot \exp(2^{2})\cdot \exp(3^{2})\cdot \ldots \cdot \exp(n^{2})=\exp \left({\frac {n(n+1)(2n+1)}{6}}\right)}