a)1+2+4+8+....+2×2ⁿ=131071
avem suma termenilor unei progresii geometrice
b₁=1
q=2
S(n+1)=b₁(1-qⁿ⁺¹)/(1-q)
1(1-2ⁿ⁺¹)/(1-2)=131071
1(1-2ⁿ⁺¹)/(-1)=131071
-1(1-2ⁿ⁺¹)=131071
2ⁿ⁺¹-1=131071
2ⁿ⁺¹=131072
2ⁿ⁺¹=2¹⁷
n+1=17
n=16
b)
2³+2²+2+....+2³⁻ⁿ=127/8
avem suma termenilor unei progresii geometrice
b₁=2³
q=1/2
b(k)=b₁×q^(k-1)
2³⁻ⁿ=2³×1/2^(k-1)
2³⁻ⁿ=2³×2^(1-k)
2³⁻ⁿ=2^(4-k)
3-n=4-k
k=n+1
S(k)=b₁×q^(k-1)/(q-1)
2³(2⁻ⁿ⁻¹-1)/(1/2-1)=127/8
-2⁴(2⁻ⁿ⁻¹-1)=127/8
-2⁷(2⁻ⁿ⁻¹-1)=127 2⁷=128
-2⁻ⁿ⁺⁶+2⁷=127
-2⁻ⁿ⁺⁶=127-128
-2⁻ⁿ⁺⁶=-1
-n+6=0
n=6
