作业帮问两道数列的题1 若四个数成等差数列,若顺次加上1,1,3,9后成等比数列,则成等比数列的4个数之和是是多少2 证明数列前n项和Sn=n(an),是等差数列但不一定是等比数列
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问两道数列的题1 若四个数成等差数列,若顺次加上1,1,3,9后成等比数列,则成等比数列的4个数之和是是多少2 证明数列前n项和Sn=n(an),是等差数列但不一定是等比数列
问两道数列的题
1 若四个数成等差数列,若顺次加上1,1,3,9后成等比数列,则成等比数列的4个数之和是是多少
2 证明数列前n项和Sn=n(an),是等差数列但不一定是等比数列
问两道数列的题1 若四个数成等差数列,若顺次加上1,1,3,9后成等比数列,则成等比数列的4个数之和是是多少2 证明数列前n项和Sn=n(an),是等差数列但不一定是等比数列
a(n)=a+(n-1)d,n=1,2,3,4.
a(2)+1=a+d+1=[a(1)+1]q=(a+1)q.d = (a+1)q-a-1 = (a+1)(q-1).
a(3)+3=a+2d+3=[a(1)+1]q^2=(a+1)q^2 = a+3+2(a+1)(q-1),
(a+1)[q^2 - 2(q-1) - 1] = (a+1)(q^2 - 2q + 1) = (a+1)(q-1)^2 = 2,
a不为-1,q不为1.d=(a+1)(q-1)=2/(q-1)不为0.
a+1 = 2/(q-1)^2,
a(4)+9=a+3d+9=[a(1)+1]q^3=(a+1)q^3 = a+9+3(a+1)(q-1) = (a+1)[3(q-1)+1] + 8,
(a+1)q^3 = 2q^3/(q-1)^2 = (a+1)[3(q-1)+1] + 8 = 2[3(q-1)+1]/(q-1)^2 + 8 = 2(3q-2)/(q-1)^2 + 8,
q^3 = (3q-2) + 4(q-1)^2 = 3q-2 + 4q^2 - 8q + 4 = 4q^2 - 5q + 2
0 = q^3 - 4q^2 + 5q - 2 = q^3 - q^2 - 3q^2 + 3q + 2q - 2 = (q-1)[q^2 - 3q + 2] = (q-1)(q-2)(q-1),
q不为1,所以,q=2.
a+1=2/(q-1)^2 = 2.a=1.
d = 2/(q-1)=2.
等比数列的4个数之和 = (a+1)[1+q+q^2+q^3] =2*[1 + 2 + 2^2 + 2^3] = 2*(2^4 - 1)/(2-1) = 30.
s(n)=na(n),
s(n+1)=(n+1)a(n+1),
a(n+1)=s(n+1)-s(n)=(n+1)a(n+1)-na(n),
0 = na(n+1)-na(n) = n[a(n+1)-a(n)],
a(n+1)=a(n) = a(n) + 0,
{a(n)}是首项为a(1),公差为0的等差数列.
另外,
a(n+1)=a(n)=...=a(1),
a(1)不为0时,a(n)=a(1)=a(1)*1^(n-1),{a(n)}是首项为a(1),公比为1的等比数列.
a(1)=0时,{a(n)=0}是常数数列,但不满足等比数列的定义[a(1)不为0,公比不为0],因此不是等比数列.