已知等差数列{an}的前n项和为Sn,且an=1/(1+2+3+…+n),求S2013的值.
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已知等差数列{an}的前n项和为Sn,且an=1/(1+2+3+…+n),求S2013的值.
S2013 = a1 + a2 + a3 + …… + a2023
= 1/1 + 1/(1+2) + 1/(1+2+3) + …… + 1/(1+2+3+……+2013)
=(1/2)* 1/(1*2) + (1/2)* 1/(2*3) + (1/2)* 1/(3*4) + …… + (1/2)* 1/(2013*2014)
= (1/2)[ 1/(1*2) + 1/(2*3) + 1/(3*4) +……+ 1/(2013*1014)]
= (1/2)*[ 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + …… + 1/2012 - 1/2014)
= (1/2)*(1 - 1/2014)
= (1/2) * (2013/2014)
= 2013/4028
= 1/1 + 1/(1+2) + 1/(1+2+3) + …… + 1/(1+2+3+……+2013)
=(1/2)* 1/(1*2) + (1/2)* 1/(2*3) + (1/2)* 1/(3*4) + …… + (1/2)* 1/(2013*2014)
= (1/2)[ 1/(1*2) + 1/(2*3) + 1/(3*4) +……+ 1/(2013*1014)]
= (1/2)*[ 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + …… + 1/2012 - 1/2014)
= (1/2)*(1 - 1/2014)
= (1/2) * (2013/2014)
= 2013/4028
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