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Let a sequence{a(n)} be defined by a(n)=...

Let a sequence`{a_(n)}` be defined by `a_(n)=(1)/(n+1)+(1)/(n+2)+(1)/(n+3)+"....."(1)/(3n)`, then

A

`a_(2)=(11)/(12)`

B

`a_(2)=(19)/(20)`

C

`a_(n+1)-a_(n)=((9n+5))/((3n+1)(3n+2)(3n+3))`

D

`a_(n+1)-a_(n)=(-2)/(3(n+1))`

Text Solution

Verified by Experts

The correct Answer is:
B, C
`:.a_(n)=(1)/(n+1)+(1)/(n+2)+(1)/(n+3)+"....."(1)/(3n)`
`a_(n)=(1)/(n+1)+(1)/(n+2)+(1)/(n+3)+"....."(1)/(n+2n)`
`a_(n)=sum_(alpha=1)^(2n)(1)/(n+alpha)`
`a_(2)=sum_(alpha=1)^(4)(1)/(2+alpha)=(1)/(3)+(1)/(4)+(1)/(5)+(1)/(6)=(20+15+12+10)/(60)`
`=(57)/(60)=(19)/(20)`
Now, `a_(n+1)-a_(n)=((1)/(n+2)+(1)/(n+3)+"......."+(1)/(3n+3))-((1)/(n+1)+(1)/(n+2)+"......+(1)/(3n))`
`=(1)/(3n+1)+(1)/(3n+2)+(1)/(3n+3)-(1)/(n+1)`
`=(1)/(3n+1)+(1)/(3n+2)-(2)/(3(n+1))`
`=(9n^(2)+15n+6+9n^(2)+12n+3-18n^(2)-18n-4)/((3n+1)+(3n+2)+(3n+3))`
`=(9n+5)/((3n+1)+(3n+2)+(3n+3))`
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