大师作文网数列题第二问
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数列题第二问
na(n+1)=(n+1)a(n)+2,
a(n+1)/(n+1) = a(n)/n + 2/[n(n+1)] = a(n)/n + 2/n - 2/(n+1),
a(n+1)/(n+1) + 2/(n+1) = a(n)/n + 2/n,
{a(n)/n + 2/n}是首项为a(1)+2=4,的常数数列.
a(n)/n + 2/n = 4,
a(n) + 2 = 4n,
a(n) = 4n - 2.
b(n)=bq^(n-1),
b=b(1)=a(2)=4*2-2=6,
bq=b(2)=a(5)=4*5-2=18.q = (bq)/b = 18/6 = 3.
b(n)=6*3^(n-1)= 2*3^n.
a(n)=4n-2=2(2n-1),
2k-1 = 3^n,k = [3^n + 1]/2.
a[(3^n + 1)/2] = 2[3^n + 1 - 1] = 2*3^n = b(n),
b(n)是{a(n)}中的第(3^n+1)/2项.
c(n)=a(n)b(n)=(4n-2)*2*3^n = (8n-4)3^n,
s(n) = c(1)+c(2)+...+c(n-1)+c(n)
=(8*1-4)3 + (8*2-4)3^2 + ...+ [8(n-1)-4]3^(n-1) + (8n-4)3^n,
3s(n) = (8*1-4)3^2 + (8*2-4)3^3 + ...+ [8(n-1)-4]3^n + (8n-4)3^(n+1),
2s(n) = 3s(n)-s(n) = -(8*1-4)3 - 3^2- ...- 3^n + (8n-4)3^(n+1)
= (8n-4)3^(n+1) - 12 - 3^2[1+3+...+3^(n-2)]
= (8n-4)3^(n+1) - 12 - 9[3^(n-1) - 1]/(3-1)
= (8n-4)3^(n+1) - 12 - (9/2)[3^(n-1) - 1],
s(n) = (4n-2)3^(n+1) - 6 - (9/4)[3^(n-1) - 1]
a(n+1)/(n+1) = a(n)/n + 2/[n(n+1)] = a(n)/n + 2/n - 2/(n+1),
a(n+1)/(n+1) + 2/(n+1) = a(n)/n + 2/n,
{a(n)/n + 2/n}是首项为a(1)+2=4,的常数数列.
a(n)/n + 2/n = 4,
a(n) + 2 = 4n,
a(n) = 4n - 2.
b(n)=bq^(n-1),
b=b(1)=a(2)=4*2-2=6,
bq=b(2)=a(5)=4*5-2=18.q = (bq)/b = 18/6 = 3.
b(n)=6*3^(n-1)= 2*3^n.
a(n)=4n-2=2(2n-1),
2k-1 = 3^n,k = [3^n + 1]/2.
a[(3^n + 1)/2] = 2[3^n + 1 - 1] = 2*3^n = b(n),
b(n)是{a(n)}中的第(3^n+1)/2项.
c(n)=a(n)b(n)=(4n-2)*2*3^n = (8n-4)3^n,
s(n) = c(1)+c(2)+...+c(n-1)+c(n)
=(8*1-4)3 + (8*2-4)3^2 + ...+ [8(n-1)-4]3^(n-1) + (8n-4)3^n,
3s(n) = (8*1-4)3^2 + (8*2-4)3^3 + ...+ [8(n-1)-4]3^n + (8n-4)3^(n+1),
2s(n) = 3s(n)-s(n) = -(8*1-4)3 - 3^2- ...- 3^n + (8n-4)3^(n+1)
= (8n-4)3^(n+1) - 12 - 3^2[1+3+...+3^(n-2)]
= (8n-4)3^(n+1) - 12 - 9[3^(n-1) - 1]/(3-1)
= (8n-4)3^(n+1) - 12 - (9/2)[3^(n-1) - 1],
s(n) = (4n-2)3^(n+1) - 6 - (9/4)[3^(n-1) - 1]