Question

Find the sum to n terms of the series :`1/(1xx2)+1/(2xx3)+1/(3xx4)+dotdotdot`

Answer 1

Correct Answer - `(n)/((n+1))`
`T_(n)=(1)/(n(n+1))=(A)/(n)+(B)/(n+1)=(A(n+1)+Bn)/(n(n+1))=((A+B)n+A)/(n(n+1)).`
`therefore A+B=0 and A=1 rArr A =1 and B=-1`
`therefore T_(n)=((1)/(n)-(1)/(n+1))`
`therefore T_(1)=((1)/(1)-(1)/(2))`
`T_(2)=((1)/(2)-(1)/(3))`
` T_(3)=((1)/(3)-(1)/(4))`
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`T_(n-1)=((1)/(n-1)-(1)/(n))`
`T_(n)=((1)/(n)-(1)/(n+1)).`
Adding columnwise, we get
`S_(n)=(T_(1)+T_(2)+T_(3)+...T_(n))=(1-(1)/(n+1))=((n+1)-1)/(n+1)=(n)/(n+1).`