求一道关于“等差数列的前n项和”证明题.已知数列{a(下标)n}是等差数列,S(下标)n是其前n项的和,求证:S(下标)6,S(下标)12-S(下标)6,S(下标)18-S(下标)12成等差数列.设k属于N*
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求一道关于“等差数列的前n项和”证明题.已知数列{a(下标)n}是等差数列,S(下标)n是其前n项的和,求证:S(下标)6,S(下标)12-S(下标)6,S(下标)18-S(下标)12成等差数列.设k属于N*
求一道关于“等差数列的前n项和”证明题.
已知数列{a(下标)n}是等差数列,S(下标)n是其前n项的和,求证:S(下标)6,S(下标)12-S(下标)6,S(下标)18-S(下标)12成等差数列.设k属于N*(正整数集),S(下标)k,S(下标)2k-S(下标)k,S(下标)3k-S(下标)2k,成等差数列.
求一道关于“等差数列的前n项和”证明题.已知数列{a(下标)n}是等差数列,S(下标)n是其前n项的和,求证:S(下标)6,S(下标)12-S(下标)6,S(下标)18-S(下标)12成等差数列.设k属于N*
已知数列{a‹n›}是等差数列,S‹n›是其前n项的和,求证:S‹6›,S₁₂-S‹6›,S‹18›-S₁₂成等差数列.设k∈N+,S‹k›,S‹2k›-S‹k›,S‹3k›-S‹2k›成等差数列.
证明:①S‹6›=6a₁+15d.(1);
S₁₂-S‹6›=12a₁+66d-(6a₁+15d)=6a₁+51d.(2);
S‹18›-S₁₂=18a₁+153d-(12a₁+66d)=6a₁+87d.(3);
(2)-(1)=36d,(3)-(2)=36d,二者相等,故是等差数列.
②S‹K›=Ka₁+k(k-1)d/2.(4)
S‹2k›-S‹k›=2ka₁+2k(2k-1)d/2-[Ka₁+k(k-1)d/2]=ka₁+(3k²-k)d/2.(5)
S‹3k›-S‹2k›=3ka₁+3k(3k-1)d/2-[2ka₁+2k(2k-1)d/2]=ka₁+(5k²-k)d/2.(6)
(5)-(4)=k²d,(6)-(5)=k²d,二这相等,故是等差数列.