Question

Fie numerele: a= 1+2+2^2+2^3+...+2^101 si b=3(1+4+4^2+...4^50) Aratati ca a=b. Va rog sa ma ajutati daca stiti.

Answer 1

S = 1+2+2^2+2^3+...+2^101 |×2
2×S = 2+2^2+2^3+2^4+...+2^102
2×S - S = 2^102 - 1
S = 2^102 - 1
a = 2^102 - 1

S = 3×(1+4+4^2+...4^50) |×4
4×S = 3×( 4 + 4^2+4^3+...+4^51)
4×S - S = 3×( 4^51 - 1)
3×S = 3 × (4^51 - 1)
S = 3 × ( 2^102 - 1) : 3
S = 2^102 - 1
b = 2^102 - 1

a = b
2^102 - 1 = 2^102 - 1