已知数列{an}a1=1
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已知数列{an}a1=1
(1)若a(n+1)=an+n求通项公式an
(2)若a(n+1)=an+2^(n-1)求通项公式an
(1)若a(n+1)=an+n求通项公式an
(2)若a(n+1)=an+2^(n-1)求通项公式an
![已知数列{an}a1=1](/uploads/image/z/19948926-30-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9Da1%3D1)
1)
a1 = 1
a2 = a1 + 1
a3 = a2 + 2
a4 = a3 + 3
……
an = a(n-1) + (n-1)
以上各式子相加,消去等式2端相同的项,最后残留
an = 1 + 1 + 2 + 3 + …… (n-1)
= 1 + [1+(n-1)]*(n-1)/2
= n*(n-1)/2 + 1
2)若a(n+1)=an+2^(n-1)求通项公式an
a1 = 1
a2 = a1 + 2^0
a3 = a2 + 2^1
a4 = a3 + 2^2
……
an= a(n-1) + 2^(n-2)
以上各式子相加,消去等式2端相同的项,最后残留
an = 1 + 2^0 + 2^1 + 2^2 + …… + 2^(n-2)
= 1 + 1*[2^(n-1)-1]/(2-1)
= 2^(n-1)
a1 = 1
a2 = a1 + 1
a3 = a2 + 2
a4 = a3 + 3
……
an = a(n-1) + (n-1)
以上各式子相加,消去等式2端相同的项,最后残留
an = 1 + 1 + 2 + 3 + …… (n-1)
= 1 + [1+(n-1)]*(n-1)/2
= n*(n-1)/2 + 1
2)若a(n+1)=an+2^(n-1)求通项公式an
a1 = 1
a2 = a1 + 2^0
a3 = a2 + 2^1
a4 = a3 + 2^2
……
an= a(n-1) + 2^(n-2)
以上各式子相加,消去等式2端相同的项,最后残留
an = 1 + 2^0 + 2^1 + 2^2 + …… + 2^(n-2)
= 1 + 1*[2^(n-1)-1]/(2-1)
= 2^(n-1)
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