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an=an-1
a20=a1+19r
r= -1
a20=3+ (-19)
a20=-22
zad2.
s6= a1 * (1-q^n/1-q)= 2 * (1-(-1)^6)/-1-(-1)= 2 * (1-6/0)=2 * (-5/0)= 2 *0=0
^ to do potęgi
3, 2, 1, 0, -1, -2
an = ?
an = a₁ + (n - 1)*r
2 - 3 = -1
1 - 2 = -1
a₁ = 3
r = - 1
an = 3 + (n - 1)*( - 1)
an = 3 - n + 1
an = - n + 4
a₂₀ = - 20 + 4
a₂₀ = - 16
Odp: Dwudziesty wyraz ciągu wynosi - 16.
Zad. 2)
Dane:
a₁ = 2
q = - 1
S₆ = ?
S₆ = a₁*(1 - q^n)/(1 - q)
S₆ = 2*[1 - ( - 1)⁶]/[1 - ( - 1)]
S₆ = 2*0/2
S₆ = 0
Odp: Suma sześciu wyrazów tego ciągu wynosi 0.
an-1=an
a1+19=ra20
r= -1
3+ (-19)=a20
-22a=20
zad2.
s6= a1 * (1-q^n/1-q)= 2
2 * (1-(-1)^6)/-1-(-1)= 2
2* (1-6/0)=2
2 * (-5/0)= 2
2*0=0