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Roanca007
a fost răspuns

dovediti ca:b) 1+3+3 la puterea a doua + 3 la puterea a treia +..+ 3 la puterea a 36=(3 la putera treizeci si sapte minus 1):2

Răspuns :

Flavistin
S=3^0+3^1+3^2+3^3+....+3^36
Inmultind cu 3 obtinem:
3·S=3^0*3+3^1*3+3^2*3+3^3*3+.....+3^36*3
3·S=3^(0+1)+3^(1+1)+3^(2+1)+3^(3+1)+......+3^(36+1)
3·S=3^1+3^2+3^3+3^4+.......+3^37
3·S=(3^0+3^1+3^2+3^3+.....+3^36)+3^37-3^0
Asadar:
3·S=S+3^37-3^0
De unde:
2·S=3^37-3^0
2·S=3^37-1
S=(3^37-1):2

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