作业帮(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和(2)由Sn=n(a1+an)/2 证明{an}是等差数列(3)证明Sk,S2k-Sk,S3k-S2k为等差数列并求其公差(
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(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和(2)由Sn=n(a1+an)/2 证明{an}是等差数列(3)证明Sk,S2k-Sk,S3k-S2k为等差数列并求其公差(
(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和
(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和
(2)由Sn=n(a1+an)/2 证明{an}是等差数列
(3)证明Sk,S2k-Sk,S3k-S2k为等差数列并求其公差
(4)求满足裂项相消的条件
(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和(1)已知数列 an 的前n项和Sn=-2n2+3n+51 求数列{|an|}的前n项和(2)由Sn=n(a1+an)/2 证明{an}是等差数列(3)证明Sk,S2k-Sk,S3k-S2k为等差数列并求其公差(
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(1) a_1=S_1=52. (1-1)
When n>=2, we have a_n=S_n-S_{n-1}=5-4n. (1-2)
Let T_n be the partial sum of the sequence |a_n|. Then T_1=a_1=52. For n>=2, we have a_n<0 and thus
T_n=a_1-(S_n-a_1...
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(1) a_1=S_1=52. (1-1)
When n>=2, we have a_n=S_n-S_{n-1}=5-4n. (1-2)
Let T_n be the partial sum of the sequence |a_n|. Then T_1=a_1=52. For n>=2, we have a_n<0 and thus
T_n=a_1-(S_n-a_1)=2n^2-3n+53. (1-3)
It is easy to check the formula (1-3) holds for n=1 too.
(2) The assumed formula does not hold because for n>=2, we have n(a_1+a_n)/2=n(57-4n)/2, different from S_n. Also, the sequence a_n is not an arithmetic progression because of a_1.
(3) The proposition holds if a_n is an arithmetic progression. However, as illustrated in (2), the sequence a_n is not arithmetic. So the proposition does not hold.
(4) The desired condition is not clear from the statement of the problem.
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(1) (2) (3) (4)不懂题目意思
