Question

高一数学数列问题

1）等比数列an中，S1=1，S8=3，则a17+a18+a19+a20的值（这个要详解+方法)
2)数列an的前n项Sn=2×(3的n次方)—2.求a1²+a2²+a3²+....+an²
3）设数列an是公差不为0的等差数列，求证：(1分之a1×a2)+(1分之a2a3)+(1分之a3a4)+....+(1分之an.an+1)=n分之 a1.an+1

Answer 1

S8=a1(1-q^8)/(1-q)
3=(1-q^8)/(1-q)
3(1-q)=1-q^8
3-3q=1-q^8
q^8-3q+2=0
(q^4-2)(q^4-1)=0
q^4=2或q^4=1
当q^4=2时
a1=s1=1
S8=a1(1-q^8)/(1-q)=a1(1-q^4)(1+q^4)/(1-q)
s8/s4=1+q^4
3/s4=1+q^4
3/s4=1+2
s4=1

a17+a18+a19+a20
=q^16*(a1+a2+a3+a4)
=(q^4)^4*s4
=2^4*1
=16

s16=a1(1-q^16)/(1-q)=a1(1-q^8)(1+q^8)/(1-q)
s16/s8=1+q^8
s16/s8=1+(q^4)^2
s16/3=1+2^2
s16=15

当q^4=1时
a1=s1=1
S8=a1(1-q^8)/(1-q)=a1(1-q^4)(1+q^4)/(1-q)
s8/s4=1+q^4
3/s4=1+q^4
3/s4=1+1
s4=3/2

a17+a18+a19+a20
=q^16*(a1+a2+a3+a4)
=(q^4)^4*s4
=3/2*1
=3/2

a1=2*3-2=4
sn=2*3^n-2
s(n-1)=2*3^(n-1)-2
an=sn-s(n-1)
=4*3^(n-1)(n>=2)
an^2=16*3^(2n-2)
a1²+a2²+a3²+....+an²
=16*3^0+16*3^2+16*3^4+......+16*3^(2n-2)
=16*[3^0+3^2+3^4+......+3^(2n-2)]
=16*1*(1-9^n)/(1-9)
=2(9^n-1)
=2*9^n-2

(1分之a1×a2)+(1分之a2a3)+(1分之a3a4)+....+(1分之an.an+1)
=1/(a2-a1)*(1/a1-1/a2)+1/(a3-a2)*(1/a2-1/a3)+1/(a4-a3)*(1/a3-1/a4)+.......+1/(an+1-an)*(1/an-1/an+1)
=1/d*[1/a1-1/a2+1/a2-1/a3+1/a3-1/a4+...............+1/an-1/a(n+1)]
=1/d[1/a1-1/a(n+1)]
=1/d[a(n+1)-a1]/[a1*a(n+1)]
=1/d*nd/[a1*a(n+1)]
=n/[a1*a(n+1)]

Answer 2

根据已知反推