数列an中,已知a1=1/2,且前n项和Sn=n^2an,则an=
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数列an中,已知a1=1/2,且前n项和Sn=n^2an,则an=
/>(1)
S2 = 2^2 * a2 = a1 + a2 = 1/2 + a2
a2 = 1/6
S3 = 3^2 * a3 = a1 + a2 + a3 = 1/2 + 1/6 + a3
a3 = 1/12
S4 = 4^2 * a4 = a1 + a2 + a3 + a4 = 1/2 + 1/6 + 1/12 + a4
a4 = 1/20
(2)
猜测{an}的通项公式是an = 1/[n(n+1)]
证:
当n = 2时,有
S2 = 2^2 * a2 = a1 + a2 = 1/2 + a2
a2 = 1/6 = 1/[2*(2+1)]
假设当n = N时,有aN = 1/[N(N+1)],SN = N^2 * aN = N/(N+1),则
当n = N+1时,有
SN+1 = (N+1)^2 * aN+1 = a1 + a2 + …… + aN + aN+1 = N/(N+1) + aN+1
aN+1 = [N/(N+1)]/[(N+1)^2 - 1] = 1/[(N+1)(N+2)]
所以
当n = N+1,公式成立
所以,对任意N,都有aN+1 = 1/[(N+1)(N+2)],该命题成立.
S2 = 2^2 * a2 = a1 + a2 = 1/2 + a2
a2 = 1/6
S3 = 3^2 * a3 = a1 + a2 + a3 = 1/2 + 1/6 + a3
a3 = 1/12
S4 = 4^2 * a4 = a1 + a2 + a3 + a4 = 1/2 + 1/6 + 1/12 + a4
a4 = 1/20
(2)
猜测{an}的通项公式是an = 1/[n(n+1)]
证:
当n = 2时,有
S2 = 2^2 * a2 = a1 + a2 = 1/2 + a2
a2 = 1/6 = 1/[2*(2+1)]
假设当n = N时,有aN = 1/[N(N+1)],SN = N^2 * aN = N/(N+1),则
当n = N+1时,有
SN+1 = (N+1)^2 * aN+1 = a1 + a2 + …… + aN + aN+1 = N/(N+1) + aN+1
aN+1 = [N/(N+1)]/[(N+1)^2 - 1] = 1/[(N+1)(N+2)]
所以
当n = N+1,公式成立
所以,对任意N,都有aN+1 = 1/[(N+1)(N+2)],该命题成立.
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