Question

1+2+3+...+100-1-3-5-...-99

Answer 1

1+2+3+...+100-1-3-5-...-99=
2+4+6+8+...+100=
2(1+2+3+...50)=
2×(50(50+1) ÷ 2)=
50×51=
2550

Answer 2

1 + 2 + 3 + ........+ 100 - 1 - 3 - 5 - .......- 99 =

= ( 1 + 2 + 3 + ......+ 100 ) - ( 1 + 3 + 5 + .....+ 99 ) =

→ 1 + 2 + 3 + ....+ 100 → suma primelor 100 de numere consecutive

→ 1 + 3 + 5 + ....+ 99 → suma numerelor impare consecutive

( 99 - 1) : 2 + 1 = 98 : 2 + 1 = 49 + 1 = 50 termeni are suma nr. impare consecutive

[ 100 × ( 1 + 100 ) / 2 ] - [ 50 × ( 1 + 99 ) / 2] =

= ( 50 × 101 ) - ( 50 × 50 ) =

= 5 050 - 2 500 =

= 2 550

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al doilea mod

1 + 2 + 3 + 4 + .........+ 98 + 99 + 100 --- ( minus )

1 + 3 +..........+ 97 + 99

__________________________

= ( 1 - 1) + 2 + ( 3 - 3 ) + 4 + ( 5 - 5 ) + ..........+ 98 + ( 99 - 99 ) + 100=

= 2 + 4 + 6 + ..................98 + 100

= 2 ( 1 + 2 + 3 + 4 + ........+ 49 + 50 ) =

= 2 × ( 50 × 51 ) / 2 =

= 2 550