Metamath Proof Explorer


Theorem fsum1

Description: The finite sum of A ( k ) from k = M to M (i.e. a sum with only one term) is B i.e. A ( M ) . (Contributed by NM, 8-Nov-2005) (Revised by Mario Carneiro, 21-Apr-2014)

Ref Expression
Hypothesis fsum1.1
|- ( k = M -> A = B )
Assertion fsum1
|- ( ( M e. ZZ /\ B e. CC ) -> sum_ k e. ( M ... M ) A = B )

Proof

Step Hyp Ref Expression
1 fsum1.1
 |-  ( k = M -> A = B )
2 fzsn
 |-  ( M e. ZZ -> ( M ... M ) = { M } )
3 2 adantr
 |-  ( ( M e. ZZ /\ B e. CC ) -> ( M ... M ) = { M } )
4 3 sumeq1d
 |-  ( ( M e. ZZ /\ B e. CC ) -> sum_ k e. ( M ... M ) A = sum_ k e. { M } A )
5 1 sumsn
 |-  ( ( M e. ZZ /\ B e. CC ) -> sum_ k e. { M } A = B )
6 4 5 eqtrd
 |-  ( ( M e. ZZ /\ B e. CC ) -> sum_ k e. ( M ... M ) A = B )