Question

determinati x din egalitatile:(1+3^2+3^4+3^6+...+3^96+3^98)*x=3+3^2+3^3+3^4+...+3^100

Answer 1

S=1+3^2+3^4+3^6+...+3^98
9S=3^2+3^4+3^6+...+3^98+3^100
9S=(1+3^2+3^4+3^6+...+3^98)+3^100-1
9S=S+3^100-1
8S=3^100-1
S=(3^100-1)/8

Q=3+3^2+3^3+3^4+...+3^100
3Q=3^2+3^3+3^4+...+3^100+3^101
3Q=(3+3^2+3^3+3^4+...+3^100)+3^101-3
3Q=Q+3^101-3
2Q=3^101-3
Q=(3^101-3)/2

Deci (1+3^2+3^4+3^6+...+3^96+3^98)*x=3+3^2+3^3+3^4+...+3^100 <=> ((3^100-1)/8)*x=(3^101-3)/2 <=> (3^100-1)*x=4(3^101-3) <=> (3^100-1)*x=12(3^100-1) <=> x=12.