数列{an} ,{bn}满足anbn = 1,an = n2 + 3n + 2,则{bn}的前十项的和为
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数列{an} ,{bn}满足anbn = 1,an = n2 + 3n + 2,则{bn}的前十项的和为
数列{an} ,{bn}满足anbn = 1,an = n2 + 3n + 2,则{bn}的前十项的和为
数列{an} ,{bn}满足anbn = 1,an = n2 + 3n + 2,则{bn}的前十项的和为
bn=1/( n2 + 3n + 2)
=1/((n+1)(n+2))
S10=1/(2*3)+1/(3*4).+1/(11*12)
=1/2-1/3+1/3-.+1/11-1/12
=1/2-1/12
=5/12
. an=(n+1)(n+2)
An与Bn互为倒数,可得出Bn通项为1/(n+1)*(n+2)
Bn=1/(n+1)*(n+2)=1/(n+1)-1/(n+2)
故有:
{B}的前十项和=(1/2-1/3)+(1/3-1/4) + (1/4-1/5)+...+(1/11-1/12)
=1/2-1/3+1/3-1/4 + 1/4-1/5+...+1/11-1/12
=1/2-1/12=5/12
anbn = 1,所以bn=1/an
an = n2 + 3n + 2
所以bn=1/n2 + 3n + 2=1/(n+1)(n+2)
{bn}的前十项的和=1/2*3+1/3*4+....+1/11*12
因为1/2*3=1/2-1/3,1/3*4=1/3-1/4,以此类推
所以{bn}的前十项的和=1/2-1/3+1/3-1/4+....+1/11-1/12
=1/2-1/12=5/12
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