设级数∑(-1)∧(n-1)Un=2,∑U(2n-1)=5则∑Un=?
举一反三
- 若级数∑n=1∞(u2n-1+u2n)收敛,则(). A: ∑n=1∞un必收敛; B: ∑n=1∞un未必收敛; C: ∑n=1∞un收敛; D: ∑n=1∞un发散·
- 设∞∑(n=1)Un为正项级数,命题如果(Un+1)/Un
- Which one of the following sequences is not covergent? A: un=∑nk=1sink2k,n=1,2,⋯. B: un=cos(1!)1⋅2+cos(2!)2⋅3+cos(3!)3⋅4+⋯+cos(n!)n⋅(n+1),n=1,2,⋯. C: un=∑nk=1(−1)k−11k,n=1,2,⋯. D: un=(1+3n(−1)n)1/n,n=1,2,⋯.
- 当$|z|<0.5$时左边序列$x[n]$为 A: $[(\frac{1}{2})^n-2^n]u[-n-1]$ B: $[(\frac{1}{2})^n+2^n]u[-n-1]$ C: $[2^n-(\frac{1}{2})^n]u[-n-1]$ D: $[2^n+(-\frac{1}{2})^n]u[-n-1]$
- 设α1,α2,…,αn是n维列向量,又A=(α1,α2,…,αn),B=(αn,α1,…,αn-1),若|A|=3,则|A+B|=______.