InterviewSolution
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InterviewSolution
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`a in R` के लिए `a ne=1,` यदि `lim_(ntooo)(1^(a)+2^(a)+।।।+n^(a))/((n+1)^(a-1)[(na+1)+(na+2)+।।।+(na+n)])=1/60` तक a का मान ज्ञात कीजिए। |
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Answer» माना, `lim_(ntooo)(1^(a)+2^(a)+...+n^(a))/((n+1)^(a-1)[(na+1)+(na+2)+...+(na+n)])` `=lim_(xtooo)(n^(a)[((1)/(n))^(a)+((2)/(n))^(a)+...((n)/(n))^(a)])/((n+1)^(a-1)[n^(2)a+(n(n+1))/(2)])` `=lim_(xtooo)(n^(a)*n[1/n underset(r=1)overset(n)sum((r)/(a))^(a)])/((n+1)^(a-1). n(na+((n+1))/(2)))` `=lim_(xtooo)(n^(a))/((n+a)^(a-1))*((1)/(a+1))/(na+(n+1)/(2))` `=lim_(ntooo)(1)/((1+(1)/(n))^(a-1))*((1)/(a+1))/(a+(1)/(2)+(1)/(2n))` `=(2)/((a+1)(2a+1))=1/60` (दिया है) `implies(a+1)(2a+1)=120` ` impliesa=7, -17/2` |
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