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FuriaNoptii
a fost răspuns

Buna. mati putea ajuta va rog la acest exercitiu? S=(3+7+11+...+103)-(1+5+9+...+101)

Răspuns :

Icecon2005
S=(3+7+11+...+103)-(1+5+9+...+101)

S = S1 - S2

S1=(3+7+11+...+103)=3+(3+1×4)+(3+2×4)+(3+1×4)+(3+3×4)+....+(3+25×4)=

S1=3+3+3+.....3 (de 25+1 ori)+4×(1+2+3+4+...+25)=
=26×3+4×(25×26):2

=78+2×25×26=78+1300=1378

S = S1 - S2 = 1378 - S2

S2=(1+5+9+...+101)
termenii sunt din 4 in 4
S2=(4×0+1)+(4×1+1)+.....(4×25+1)=
S2=4×0+4×1+4×2+......+4×25+1+1+1+1......+1(de 26 de ori)
S2=4(0+1+2+3+......+25)+26×1
S2=4×[(25×26):2]+26
S2=1326

S = S1 - S2 = 1378 - 1326 = 52

S = 52

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