Question

Rozwiąż równanie 1+3+5+...+x = 81 to jest zadanie z ciągów

Answer 1

1+3+5+...+x = 81

 

1, 3, 5, ...., x

a1 = 1

r = a2 - a1 = 3 - 1 = 2

an = x

 

an = a1 + (n - 1) * r

x = 1 + (n - 1) * 2

x = 1 + 2n - 2

x = 2n - 1

x + 1 = 2n

n = (x  +1) / 2

 

Sn = [(a1 + an) * n ] / 2 = [ (1 + x) * (x  +1)/2 ]/2

81 = [ (1 + x) * (x  +1)/2 ]/2

162 =  (1 + x) * (x  +1)/2

324 = (x  +1)(x  +1)

x² + 2x  +1 = 324

x² + 2x - 323 = 0

Δ = 4 + 1292 =

√Δ = 36

x1 = (-2 - 36) / 2 = -38/2 = -19

x2 = (-2 + 36) / 2 = 34/2 = 17

 

n = (x  +1) / 2 = (-19 + 1) / 2 = -9 ∉ D

n = (x  +1) / 2 = (17 + 1) / 2 = 9 ∈ D

odp. x = 17

Answer 2

a₁=1

a₂=3

a₃=5

r=3-1=2

an=x

Sn=81

Sn=2a₁+(n-1)r/2   ×n

81=[2×1+(n-1)×2 ] /2  ×n

162=[2+2n-2]n

162=2n²

n²=162;2

n²=81

n=9 lub n=-9, ale skoro n∈N, więc n=9

a₉=x

a₉=a₁+8r

x=1+8×2=17