lim(ntooo) (((n+1)^(1//3))/(n^(4//2))+((...

`lim_(ntooo) (((n+1)^(1//3))/(n^(4//2))+((n+2)^(1//3))/(n^(4//3))+. . . . +((2n)^(1//3))/(n^(4//3)))` is equal to

`(4)/(3)(2)^(4//3)`

`(3)/(4)(2)^(4//3) - (4)/(3)`

`(3)/(4)(2)^(4//3) - (3)/(4)`

`(3)/(4)(2)^(3//4)`

Text Solution

Verified by Experts

The correct Answer is:

Let `p=underset(ntooo)(lim)(((n+1)^(1//3))/(n^(4//3))+((n+2)^(1//2))/(n^(4//3))+ . . . + ((2n)^(1//3))/(n^(4//3)))`
`=underset(n tooo)(lim)sum_(r=1)^(n)((n+r)^(1//3))/(n^(4//3))`
`=underset(ntooo)(lim)sum_(r=1)^(n)((1+(r)/(n))^(1//3)n^(1//3))/(n^(4//3))`
`=underset(n to oo)(lim)(1)/(n)sum _(r=1)^(n)(1+(r)/(n))^(1//3)`
Now , as per integration as limit of sum.
Let `(r)/(n) =x and (1)/(n)=dx` `[:' n to oo]`
Then , upper limit of integral is 1 and lower limit of integral is 0.
So , `p=int_(0)^(1)(1+x)^(1//3)dx[ :' underset(n tooo)(lim)(1)/(n)sum_(r=1)^(n)f ((r)/(n))= int_(0)^(x) f (x) dx ]`
`=[ (3)/(4)(1+x)^(4//3)]_(0)^(1)=(3)/(4)(2^(4//3)-1)=(3)/(4)(2)^(4//3)-(3)/(4)`

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lim_(n to oo) (3^(n)+4^(n))^(1//n) is equal to

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`(1)/(2)log 2`

`(1)/(3)log 2`

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`log 2`