作业帮an是等比数列,a1+a2+a3=18,a2+a3+a4=-9,lim(a1+a2+.+an)
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an是等比数列,a1+a2+a3=18,a2+a3+a4=-9,lim(a1+a2+.+an)
an是等比数列,a1+a2+a3=18,a2+a3+a4=-9,lim(a1+a2+.+an)
an是等比数列,a1+a2+a3=18,a2+a3+a4=-9,lim(a1+a2+.+an)
a1+a2+a3=18
a1(1+q+q^2)=18
a2+a3+a4=-9
a1q(1+q+q^2)=-9
相除,q=-1/2
a1=18/(1+q+q^2)=24
a1+a2+……+an=24*[1-(-1/2)^n]/(1+1/2)
n→∞
(-1/2)^n→0
所以极限=24*1/(3/2)=16
a(n) = aq^(n-1), n = 1,2,...
18 = a(1) + a(2) + a(3) = a + aq + aq^2,
-9 = a(2) + a(3) + a(4) = aq + aq^2 + aq^3 = q[a + aq + aq^2] = 18q,
q = -1/2.
18 = a[1 - 1/2 + 1/4] = 3a/4,
a...
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a(n) = aq^(n-1), n = 1,2,...
18 = a(1) + a(2) + a(3) = a + aq + aq^2,
-9 = a(2) + a(3) + a(4) = aq + aq^2 + aq^3 = q[a + aq + aq^2] = 18q,
q = -1/2.
18 = a[1 - 1/2 + 1/4] = 3a/4,
a = 24.
lim(n->正无穷)[a(1)+a(2)+...+a(n)] = lim(n->正无穷)[a(1-q^n)/(1-q)]
= lim(n->正无穷)24[1-(-1/2)^n]/(1+1/2) = 16 - lim(n->正无穷)(-1/2)^n
= 16.
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