等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),...等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),且b1=3,求[1
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![等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),...等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),且b1=3,求[1](/uploads/image/z/1549993-49-3.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94S5%3D45%2CS6%3D60%3F%E2%91%A0%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%3F%E2%91%A1%E8%8B%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3bn%2B1-bn%EF%BC%9Dan%28n%E2%88%88N%2A%EF%BC%89%2C...%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94S5%3D45%2CS6%3D60%3F%E2%91%A0%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%3F%E2%91%A1%E8%8B%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3bn%2B1-bn%EF%BC%9Dan%28n%E2%88%88N%2A%EF%BC%89%2C%E4%B8%94b1%3D3%2C%E6%B1%82%5B1)
等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),...等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),且b1=3,求[1
等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),...
等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),且b1=3,求[1/bn]的前n项和Tn
等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),...等差数列an的前n项和为Sn,且S5=45,S6=60?①求an的通项公式an?②若数列an满足bn+1-bn=an(n∈N*),且b1=3,求[1
①S5=45,S6=60
a6=S6-S5=15
S6=(a1+a6)*3
a1=5,d=2,an=2n+3
②bn+1-bn=an=2n+3
b2-b1=5
b3-b2=7
...
bn-bn-1=2n+1,叠加法
bn=n^2+2n,1/bn=1/n(n+2)=(1/2)[(1/n)-(1/n+2)
然后就能算Tn了
S5=5*a1+5*4*d/2=45 S6=6*a1+6*5*d/2=60 解方程组得a1,d
b(n+1)-bn=an可展开,累加法求解。求出bn后,再用列项相消法求解Tn.我手头没笔,你百度一下“数列求和”你会有大收获。
an=2n+3
bn=2n*n-3+n