For ` a in R` ( the set of all real numbers) , `a ne-1,lim_(ntooo) ((1^(a)+2^(a) + . . . +n^(a)))/((n+1)^(a-1)[(na+1)+(na+2) + . . . +(na+n)])=(1)/(60)`. Then, `a` is equal to

Options

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

Answer

B, D

Explanation

PLAN Converting Infinite series into definite Integral I .e. `underset(n tooo)(lim)(h(n))/(n)`
` underset( n tooo)(lim)sum_(r =g(n))^(h(n))f ((r)/(n))= intf (x)dx`
`underset(n tooo)(lim) (g(n))/(n)` where , `(r)/(n)` replaced with integral.
Here, `underset(n tooo)(lim)(1^(a)+2^(a)+ . . . +n^(a))/((n+1)^(a^(-1)){(na+1)+(na+2)+ . . . +(na+n)})=(1)/(60)`
`rArrunderset(ntooo)(lim)(sum_(n =1)^(n)r^(a))/((n +1)^(a-1)*[n^(2)a+(n(n+1))/(2)])=(1)/(60)`
`rArr underset(n tooo)(lim)(2sum_(r=1)^(n)((r)/(n))^(a))/((1+(1)/(n))^(a-1)*(2na+n+1))=(1)/(60)`
`rArr underset(n tooo)(lim)(1)/(n)(2sum_(r=1)^(n)((r)/(n))^(a))underset(ntoo)(lim)(1)/((1+(1)/(n))^(a-1)(2a+1+(1)/(n)))=(1)/(60)`
`rArr 2 int_(0)^(1)(x^(a))dx*(1)/(1*(2a+1))=(1)/(60)`
`rArr (2*[x^(a+1)]_(0)^(1))/((2a+1)*(a+1))=(1)/(60)`
`:. (2)/((2a+1)(a+1))=(1)/(60)rArr(2a+1)(a+1)=120`
`rArr 2a^(2)+3a+1-120=0rArr2a^(2)+3a-119=0`
`rArr (2a +17)(a-7)=0rArra=7, (-17)/(2)`