在数列{an}中,若a1=1,an+1=an+2(n≥1),则该数列的通项an=______
A: 2n
B: 2n-1
C: 2n+1
D: 2n-2
E: 2n-3
A: 2n
B: 2n-1
C: 2n+1
D: 2n-2
E: 2n-3
举一反三
- 已知数列{an}满足,a1=1,a2=2,an+2=an+an+12,n∈N×.
- 设`\n`阶方阵`\A`满足`\|A| = 2`,则`\|A^TA| = ,|A^{ - 1}| = ,| A^ ** | = ,| (A^ ** )^ ** | = ,|(A^ ** )^{ - 1} + A| = ,| A^{ - 1}(A^ ** + A^{ - 1})A| = `分别等于( ) A: \[4,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] B: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n + 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] C: \[4,\frac{1}{2},{2^{n + 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\] D: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\]
- (本小题满分16分)设数列{an}满足:a1=1,a2=2,an+2=(n≥1,n∈N*).
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