Question

If $\underset{n\to \infty }{lim}\frac{1^a+2^a+.....+n^a}{(n+1{)}^{a-1}[(na+1)+(na+2)+.....+(na+n)]}=\frac{1}{60}$ for some positive real number

a, then a is equal to.

• A. $\frac{15}{2}$
• B. $8$
• C. $\frac{17}{2}$
• D. $7$

Answer 1

The correct option is D $7$

$\underset{n\to \infty }{lim}\frac{1^a+2^a+3^a+...+n^a}{(n+1{)}^{a-1}[(na+1)+(na+2)+...+(na+n)]}=\frac{1}{60}$

$\underset{n\to \infty }{lim}\frac{1^a+2^a+....+n^a}{(n+1{)}^{a-1}[n^{2}a+\frac{n(n+1)}{2}]}=\frac{1}{60}$

$\underset{n\to \infty }{lim}\frac{2\left[\sum _{k=1}^n{k}^a\right]}{(n+1{)}^{a-1}[2n^{2}a+n^2+n]}=\frac{1}{60}$

$\underset{n\to \infty }{lim}\frac{2\sum _{k=1}^n{\left(\frac{k}{n}\right)}^a}{\frac{(n+1{)}^{a-1}}{n^{a-1}}\left[\frac{2n^{2}a+n^2+n}{n}\right]}=\frac{1}{60}$

$\underset{n\to \infty }{lim}\frac{2\sum _{k=1}^n{\left(\frac{k}{n}\right)}^a}{{(1+\frac{1}{n})}^{a-1}[2na+n+1]}=\frac{1}{60}$

$\underset{n\to \infty }{lim}\frac{\frac{2}{n}\sum _{k=1}^n{\left(\frac{k}{n}\right)}^a}{{(1+\frac{1}{n})}^{a-1}[2a+1+\frac{1}{n}]}=\frac{1}{60}$

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^n{\left(\frac{k}{n}\right)}^a={\int }_0^1{x}^{a}dx$ [Derivative as limit of a sum]

$\therefore \frac{2{\int }_0^1{x}^{a}dx}{2a+1}=\frac{1}{60}$

$\frac{2|\left(x^{a+1}\right){|}_0^1}{(a+1)(2a+1)}=\frac{1}{60}\Rightarrow \frac{2}{(a+1)(2a+1)}=\frac{1}{60}$

$(a+1)(2a+1)=120\Rightarrow 2a^2+3a+1=120\Rightarrow 2a^2+3a-119=0$

$a=\frac{-3\pm \sqrt{9+952}}{4}=\frac{-3\pm \sqrt{961}}{4}=\frac{-3\pm 31}{4}$ since $a>o$

$\therefore \frac{-3+31}{4}=\frac{31-3}{4}=\frac{28}{4}=7$

$\therefore a=7$