For `a epsilonR` (the set of all real numbers) `a!=-1, lim_(n to oo) ((1^(a)+2^(a)+………+n^(a)))/((n+1)^(a-1)[(na+1)+(n+2)+………..+(na+n)])=1/60`
Then `a=`

Options

A. 5,7
B. `7,(17)/(2)`
C. `-(15)/(2),-(17)/(2)`
D. `7,(15)/(2)`

Answer

B

Explanation

`underset(n to oo)lim((1^(a)+2^(a)+3^(a)+....+n^(a)))/((n+1)^(a-1){(na+1)+( na+2)+...+(na+n)})=(1)/(60)`
`underset(n to oo)lim((1^(a)+2^(a)+3^(a)+....+n^(a)))/((n+1)^(a-1){n^(2)a+(n(n+1))/(2)))=(1)/(60)`
` rArr lim_(n tooo)(2(underset(r=1)overset(n)(sum)r^(a)))/(n^(a+1)(1+(1)/(n))^(a-1)(2a+1+(1)/(n)))=(1)/(60)`
`rArr2{ lim_(n tooo)((1)/(n)underset(r=1)overset(n)(sum)((r)/(n))^(2))/(n^(a+1)(1+(1)/(n))^(a-1)(2a+1+(1)/(n)))}=(1)/(60)`
`rArr2{ lim_(n tooo)((1)/(n)underset(r=1)overset(n)(sum)((r)/(n))^(2))/((2a+1))}=(1)/(60)`
`rArr (1)/(2a+1)lim_(n to oo){(1)/(n)overset(n)underset(r=1)sum((r)/(n))^(a)}=(1)/(60)`
`rArr (1)/(2a+1)(underset(0)overset(1)intx^(a)dx)=(1)/(60)`
`rArr (2)/(2a+1)[ (x^(a+1))/(a+1)]_(0)^(1)`
`rArr (2)/(2a+1)((1)/(a+1))=(1)/(60)`
`rArr 2a^(2)+3a+1=120`
`rArr 2a^(2)+3a-119=0`
`rArr 2a^(2)+17a-14a-119=0`
`rArr (2a+17)(a-7b)=0`
`rArr a=7,-(17)/(2)`