Question

(2+4+...+20)-(1+3+5+...+19) va rog mult acum imi trebuie sunt la test

Answer 1

( 2 + 4 + 6 + ...... + 20 ) - ( 1 + 3 + 5 + ..... + 19 ) = 10

= 10 × ( 2 + 20 ) / 2 - [ 10 x ( 1 + 19 ) / 2 ] =

= 5 × 22 - 5 × 20 =

= 5 × ( 22 - 20 ) =

= 5 × 2 =

= 10

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2 + 4 + 6 + ....... + 20 = 10 × 22 / 2 = 220 / 2 = 110

→ stabilim cati termeni are suma, cu ratia 2 ( 4 - 2 = 2;   6 - 4 = 2 .... )

( 20 - 2 ) : 2 + 1 = 18 : 2 + 1 = 9 + 1 = 10 termeni are suma

→ aplic formula sumei lui Gauss

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1 + 3 + 5 + ........ + 19 = 10 × 20 / 2 = 200 / 2 = 100

( 19 - 1 ) : 2 + 1 = 18 : 2 + 1 = 10 termeni are suma

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110 - 100 = 10 → diferenta dintre suma primelor 10 numere pare nenule consecutive si suma primelor 10 numere impare consecutive