Question

wyznacz a1,a3,a6,an-1,an+3,a2k+1 jwsli wyraz ogólny ciagu ma postac
a) an=3n-1
b)an=a/n+1
c)an=(-2) do potegi n *n
d) an=4
e) an=n!
f) an=n2+2n+1/3
g)an=3*2 do potegi n minus 3
h) an=√n+3,
i)an=3-n/n+2

Answer 1

Wyznacz a1,a3,a6,an-1,an+3,a2k+1 jwsli wyraz ogólny ciagu ma postac
a) an=3n-1
a1=3*1-1=3-1=2
a3=3*3-1=9-1=8
a6=3*6-1=18-1=17
an-1=3*(n-1)-1=3n-3-1=3n-4
an+3=3*(n+3)-1=3n+9-1=3n+8
a2k+1=3*(2k+1)-1=6k+3-1=6k+2
b)an=a/(n+1)
a1=a/(1+1)=a/2
a3=a/(3+1)=a/4
a6=a/(6+1)=a/7
an-1=a/(n-1)
an+3=a/(n+3)
a2k+1=a/(2k+1)
lub jeśli an=(a/n)+1
a1=(a/1)+1=a+1
a3=(a/3)+1=(a/3)+(3/3)=(a+3)/3
a6=(a/6)+1=(a/6)+(6/6)=(a+6)/6
an-1=(a/(n-1))+1=(a/(n-1))+((n-1)/(n-1))=[a+n-1]/[n-1]
an+3=(a/(n+3))+1=(a/(n+3))+((n=3)/(n+3))=[a+n+3]/[n+3]
a2k+1=(a/(2k+1))+1=(a/(2k+1))+((2k+1)/(2k+1))=[a+2k+1]/[2k+1]
c)an=(-2) do potegi n *n
a1=(-2)¹*1=-2
a3=(-2)³*3=-8*3=-24
a6=(-2)⁶*6=64*6=384
an-1=(-2) do potegi (n-1) *(n-1)=[(-2)⁻¹*(-2) do potegi n ]*(n-1)=[(-1/2)*(-2) do potegi n]*(n-1)=(-1/2)n*(-2) do potegi n+(1/2)*(-2) do potegi n
an+3=(-2) do potegi (n+3) *(n+3)=[(-2)³*(-2) do potegi n ]*(n+3)=[(-8)*(-2) do potegi n]*(n+3)=(-8)n*(-2) do potegi n-24*(-2) do potegi n
a2k+1=(-2) do potegi (2k+1) *(2k+1)=[(-2)¹(-2)² do potegi n ]*(2k+1)=[(-2)*4 do potegi k]*(2k+1)=(-4k)*4 do potegi k+(-2)*4 do potegi k
d) an=4
a1=4
a3=4
a6=4
an-1=4
an+3=4
a2k+1=4
e) an=n!
a1=1!=1
a3=3!=1*2*3=6
a6=6!=1*2*3*4*5*6=720
an-1=(n-1)!
an+3=(n+3)!=n!(n+1)(n+2)(n+3)
a2k+1=(2k+1)!=(2k)!(2k+1)
f) an=n²+2n+1/3
a1=1²+2*1+1/3=1+2+1/3=3 i 1/3
a3=3²+2*3+1/3=9+6+1/3=15 i 1/3
a6=6²+2*6+1/3=36+12+1/3=48 i 1/3
an-1=(n-1)²+2(n-1)+1/3=n²-2n+1+2n-2+1/3=n²-2/3
an+3=(n+3)²+2(n+3)+1/3=n²+6n+9+2n+6+1/3=n²+8n+15+1/3
a2k+1=(2k+1)²+2(2k+1)+1/3=4k²+4k+1+4k+2+1/3=4k²+8k+3+1/3
g)an=3*2 do potegi (n-3)
a1=3*2¹⁻³=3*2⁻²=3*(1/2)²=3*(1/4)=3/4
a3=3*2³⁻³=3*2⁰=3*1=3
a6=3*2⁶⁻³=3*2³=3*8=24
an-1=3*2 do potegi (n-1-3)=3*2 do potegi (n-4)
an+3=3*2 do potegi (n+3-3)=3*2 do potegi n
a2k=1=3*2 do potegi (2k+1-3)=3*2 do potegi (2k-2)=3*2 do potegi 2k * 2 do potęgi (-2)=3*4 do potegi k * (1/4)=(3/4)*4 do potegi k
h) an=(√n)+3
a1=√1+3=1+3=4
a3=√3+3=3+√3
a6=√6+3=3+√6
an-1=√(n-1)+3=3+√(n-1)
an+3=√(n+3)+3=3+√(n+3)
a2k+1=√(2k+1)+3=3+√(2k+1)
lub jeśli an=√(n+3)
a1=√(1+3)=4=2
a3=√(3+3)=√6
a6=√(3+6)=√9=3
an-1=√(n-1+3)=√(n+2)
an+3=√(n+3+3)=√(n+6)
a2k+1=√(2k+1+3)=√(2k+4)=√[2(k+2)]=√2*√(k+2)
i)an=3-n/(n+2)
a1=3-1/(1+2)=3-1/3=2 i 2/3
a3=3-3/(3+2)=3-3/5=2 i 2/5
a6=3-6/(6+2)=3-6/8=2 i 2/8=2 i 1/4
an-1=3-(n-1)/(n-1+2)=3-(n-1)/(n+1)=(3n+3)/(n+1)-(n-1)/(n+1)=(2n+4)/(n+1)
an+3=3-(n+3)/(n+3+2)=3-(n+3)/(n+5)=(3n+15)/(n+5)-(n+3)/(n+5)=(2n+12)/(n+1)
a2k+1=3-(2k+1)/(2k+1+2)=3-(2k+1)/(2k+3)=(6k+9)/(2k+3)-(2k+1)/(2k+3)=(4k+8)/(2k+3)