1.

निम्न श्रेणी के nपदों तथा अनन्त पदों का योग ज्ञात कीजिए | `1+(1)/(1+2)+(1)/(1+2+3)+…`

Answer» Correct Answer - `(2n)/(n+1),2`
(i) दी गई श्रेणी का n वाँ पद `T_(n)=(1)/(1+2+3+..n)=(1)/(Sigma2)=(1)/((1)/(2)n(n+1))`
`=(2)/(n(n+1))=2((1)/(n)-(1)/(n+1))`
अब `" "S_(n)=T_(1)+T_(2)+T_(3)+...+T_(n)`
`=2[{1-(1)/(2)}+{(1)/(2)-(1)/(3)}+{(1)/(3)-(1)/(4)}+...{(1)/(n)-(1)/(n+1)}]`
`=2[1-(1)/(n+1)]=2[(n+1-1)/(n+1)]=(2n)/(n+1)`
`:. ""S_(oo)=underset(n to oo)lim S_(n)[2{1-(1)/(n+1)}]=2(1-(1)/(oo))=2`


Discussion

No Comment Found

Related InterviewSolutions