问题补充:
在数列{an}中,a1=2,an+1=an+2n,则an=
答案:
∵在数列{an}中,a1=2,an+1=an+2n,
∴a2-a1=2,
a3-a2=4,
a4-a3=6,
…an+1-an=2n,
∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)
=2+2+4+6+…+2(n-1)
=2+2×(1+2+3+…+n-1)
=2+n(n-1)=n2-n+2.
∴a
时间:2021-11-04 05:27:29
在数列{an}中,a1=2,an+1=an+2n,则an=
∵在数列{an}中,a1=2,an+1=an+2n,
∴a2-a1=2,
a3-a2=4,
a4-a3=6,
…an+1-an=2n,
∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)
=2+2+4+6+…+2(n-1)
=2+2×(1+2+3+…+n-1)
=2+n(n-1)=n2-n+2.
∴a
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