已知数列{an}中,a1=1,且a*2k=a*(2k-1)+(-1)*k,a*(2k+1)=a*2k+3*k已知数列{an},a1=1,a2k=a(2k-1)+(-1)^k,a(2k+1)=a2k+3k,k=1,2,3,...(1)求a3,a5 (2)求数列an的通项公式
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![已知数列{an}中,a1=1,且a*2k=a*(2k-1)+(-1)*k,a*(2k+1)=a*2k+3*k已知数列{an},a1=1,a2k=a(2k-1)+(-1)^k,a(2k+1)=a2k+3k,k=1,2,3,...(1)求a3,a5 (2)求数列an的通项公式](/uploads/image/z/6796242-18-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1%2C%E4%B8%94a%2A2k%3Da%2A%282k-1%29%2B%28-1%29%2Ak%2Ca%2A%282k%2B1%29%3Da%2A2k%2B3%2Ak%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%2Ca1%3D1%2Ca2k%3Da%282k-1%29%2B%EF%BC%88-1%29%5Ek%2Ca%282k%2B1%29%3Da2k%2B3k%2Ck%3D1%2C2%2C3%2C...%281%29%E6%B1%82a3%2Ca5+%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
已知数列{an}中,a1=1,且a*2k=a*(2k-1)+(-1)*k,a*(2k+1)=a*2k+3*k已知数列{an},a1=1,a2k=a(2k-1)+(-1)^k,a(2k+1)=a2k+3k,k=1,2,3,...(1)求a3,a5 (2)求数列an的通项公式
已知数列{an}中,a1=1,且a*2k=a*(2k-1)+(-1)*k,a*(2k+1)=a*2k+3*k
已知数列{an},a1=1,a2k=a(2k-1)+(-1)^k,a(2k+1)=a2k+3k,k=1,2,3,...(1)求a3,a5 (2)求数列an的通项公式
已知数列{an}中,a1=1,且a*2k=a*(2k-1)+(-1)*k,a*(2k+1)=a*2k+3*k已知数列{an},a1=1,a2k=a(2k-1)+(-1)^k,a(2k+1)=a2k+3k,k=1,2,3,...(1)求a3,a5 (2)求数列an的通项公式
S1 + A1 = 2 * A1 = 3
A1 = 3/2
N> = 2时,
SN +一= 2N +1
S(n-1个)+第(n-1)= 2n-1个
2AN =(正 - 1)2
2(-2)=(-1)-2
α1-2 = -1 / 2
-2 = -1 / 2的n次方
= 2-1/2 ^ n
你能直接上图吗 ? 这种题目一般都是压抽题 题目不清楚解题是非常困难
1)a2=a1+(-1)=0
a3=a2+3=3
a4=a3+1=4
a5=a4+3*2=10
2) 因为a(2k-1)=a(2k-2)+3(k-1)
所以a2k=a(2k-1)+(-1)^k=a(2k-2)+3(k-1)+(-1)^k
a4=a2...
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你能直接上图吗 ? 这种题目一般都是压抽题 题目不清楚解题是非常困难
1)a2=a1+(-1)=0
a3=a2+3=3
a4=a3+1=4
a5=a4+3*2=10
2) 因为a(2k-1)=a(2k-2)+3(k-1)
所以a2k=a(2k-1)+(-1)^k=a(2k-2)+3(k-1)+(-1)^k
a4=a2+3*1+(-1)^2
a6=a4+3*2+(-1)^3
...
a2k=a(2k-2)+3(k-1)+(-1)^k
以上各式相加得:a2k=a2+3(k-1)lk/2+(-1)^2[1-(-1)^(k-1)]/2=3k(k-1)/2+[1+(-1)^k]/2
因此有:a(2k+1)=a2k+3k=3k(k+1)/2+[1+(-1)^k]/2
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