Question

http://i3.minus.com/jPwLpw8HTXe6n.PNG

Answer 1

i think there is no closed form !!! maybe?

Answer 2

This is complicated
http://en.wikipedia.org/wiki/Faulhaber's_formula

Answer 3

even if use LH,on solving numerator i get,log(n!)

Answer 4

post the original question ...
http://i3.minus.com/jPwLpw8HTXe6n.PNG

Answer 5

done already :P

Answer 6

am I right with the numerator derivative?

Answer 7

and on simplifying denominator..(didnt take derivative yet)
\[\Large (n+1)^{a-1} (((na)^n+\frac{n(n+1)}{2}))\]

Answer 8

\[ \Large (n+1)^{a-1} (((n^2a)+\frac{n(n+1)}{2}))\]

Answer 9

??

Answer 10

n times na = n^2 a not (na)^n

Answer 11

what next?

Answer 12

make use of Faulhaber's formula ...

Answer 13

alternates?
cause saw it first time...

Answer 14

no ... just take first term. Can't believe High Schooler get's problem like this. I myself haven't proved this formula.
\[ \frac{1}{a+1} \cdot \frac{n^{a+n}}{n^{a-1} \cdot n^2 (a+1/2)} = 1/60 \]

Answer 15

*a+1 at the top.

Answer 16

hmm..im sure there must be an alternate though,these formulaes are hard to memorize and difficult to apply..

Answer 17

still this is not working

Answer 18

bang !

Answer 19

numerically I found a=7
Limit[Sum[k^7, {k, 1, n}]/((n)^(6) (n^2 7 + n (n + 1)/2)) ,
n -> Infinity]

Answer 20

use eq.editor!

Answer 21

that is mathematica code

Answer 22

oh hmm

Answer 23

Damn ... I had been doing opposite!! putting 60 instead of 1/60

http://www.wolframalpha.com/input/?i=1%2F%28a%2B1%29*1%2F%28a%2B1%2F2%29+%3D+1%2F60

Answer 24

Unfortunately, both B) and D) works ... where the hell did you get this question?

Answer 25

i told you about IIT right :O

Answer 26

For D = -17/2, the sum does not converge .. so your answer is B), a=7

Answer 27

hmm..thanks,but how :O

Answer 28

Solve this equation
\[ \Large \lim_{n \to \infty } \frac{1}{a+1} \cdot \frac{n^{a+1}}{(n+1)^{a-1} \cdot n^2 (a+1/2)} = 1/60\]

Answer 29

u used that formula?

Answer 30

yes

Answer 31

hmm..okaythanks!

Answer 32

just remember this
\[ \sum_{k=0}^n k^a = \frac{1}{a+1 }O(n^{a+1})\]
I think you can prove this using Euler summation formula ... but let's not get there.
\[ \sum_{k=0}^n k^a \approx \int_0^n x^{k}dx\]

Answer 33

okay..!

Answer 34

let's see if anyone can come up with better method.

Answer 35

@hartnn