Question

A ) (1+2+3+...+99) : 50
b) (1+4+7+...+301) : 101
URGENT PLS

Answer 1

La A) am folosit 1+2+3+...+n=n(n+1)/2 si la B) numerele sunt din 3 in 3 asa ca pe fiecare l-am scris 3·n+1 si tot la suma lui gauss s-a ajuns(cea de la punctul A) )

Answer 2

a)   ( 1 + 2 + 3 + .... + 99 ) : 50 =

= ( 99 × 100 : 2 ) : 50 =

= 99 × 50 : 50 =

= 99

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1 + 2 + 3 + ...... + 99  → aplic formula sumei lui Gauss

= 99 termeni × ( 1 + 99 ) : 2 = 99 × 100 : 2 = 9 900 : 2 = 4950

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4 950 : 50 = 99

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b)   ( 1 + 4 + 7 + ....... + 301 ) : 101 = 151

= 101 × 151 : 101 =   ( 101 × 151 × 1 / 101 )

= 151

1 + 4 + 7 + ....... + 301 =

→ stabilesc cati termeni are suma cu ratia = 3 ( 4 - 1 = 3;   7 - 4 = 3;..... )

( 301 - 1 ) : 3 + 1 = 300 : 3 + 1 = 100 + 1 = 101 termeni are suma

→ aplic formula sumei lui Gauss

1 + 4 + 7 + ....... + 301 =  101 × ( 1 + 301 ) : 2 = 101 × 302 : 2 = 101 × 151