Question

数列2²+1/2²-1,3²+1/3²-1……(n+1)²+1/(n-1)²-1的前10项和为____

Answer 1

考察一般项：
[(n+1)²+1]/[(n+1)²-1]
=(n²+2n+2)/(n²+2n)
=1 +2/[n(n+2)]
=1+ 1/n -1/(n+2)
前10项和=1+1/1-1/3+1+1/2-1/4+...+1+1/10-1/12
=10+(1+1/2+...+1/10)-(1/3+1/4+...+1/12)
=10+1+1/2-1/11-1/12
=1495/132
前n项和=n+(1+1/2+...+1/n)-[1/3+1/4+...+1/(n+2)]
=n+ 3/2 -1/(n+1)- 1/(n+2)

Answer 2

=1+2/(2^2-1)+1+2/(3^2-1)+...+1+2/(11^2-1)
=10+2[1/(2^2-1)+1/(3^2-1)+...+1/(11^2-1)]
=10+2{[1/(2-1)-1/(2+1)]/2+[1/(3-1)-1/(3+1)]/2+...+[1/(11-1)-1/(11+1)]/2}
=10+(1/1-1/3+1/2-1/4+...+1/10-1/12)
=10+(1+1/2-1/11-1/12)
=11+1/2-1/11-1/12
=11+1/2-23/132
=11+43/132