z.1 a) 1 +2 + 3 + ... + 100 = [ 100*101]/2 = 50 *101 = 5050 Korzystałem z wzoru na sumę kolejnych liczb naturalnych: S = 1 + 2 + 3 + ... + n = [ n*(n +1)]/2 II dposób, Mamy ciag arytmetyczny: a1 = 1 oraz r = 1 an = a100 = 100 S100 = 0,5*[a1 +a100]*100 =50*[1 + 100] = 50 * 101 = 5050 S100 = 5050 =============== b) 2 + 4 + 6 + ... + 248 = Mamy ciag arytmetyczny" a1 = 2 oraz r = 2 an = 248 ale an = a1 + (n-1)*r = 2 + (n-1)*2 = 2 + 2n - 2 = 2n zatem 2n = 248 ---> n = 124 S124 = 0,5*[ a1 + a124]*124 = 62*[2 + 248] = 62*250 = 15 500 Odp. S124 = 15 500 ======================== z.2 a1 = 3; q = - 2 Mamy a4 = a1*q^3 = 3*(-2)^3 = 3*(-8) =- 24 ================================= S5 = a1*[1 - q^5]/[1 - q] = 3*[ 1 - (-2)^5]/[ 1 - (-2)] = 3*[1 + 32]/[3] = 33 S5 = 33 =============
Zad. 1 a) 1+2+3+...+100 a₁=1 a₁₀₀=100 r=1 [latex]S_{100}=frac{a_{1}+a_{100}}{2} cdot 100\ S_{100}=frac{1+100}{2} cdot 100=5050[/latex] b) 2+4+6+...+248 a₁=2 r=2 [latex]a_{n}[/latex]=2n, więc a₁₂₄=248 [latex]S_{124}=frac{a_{1}+a_{124}}{2} cdot 124\ S_{124}=frac{2+248}{2}cdot 124=15 500 [/latex] Zad. 2 a₁=3 q=-2 a₄=a₁ · q³ a₄=3 · (-8) a₄=-24 [latex]S_{5}=3 cdot frac{1-(-2)^{5}}{1-(-2)}=3 cdot frac{33}{3}=33[/latex]
