Here's a challenge everyone --- please help

1. Consider, instead, the following sum:
S=e^{ix}+2e^{i2x}+3e^{i3x}+\ldots + n e^{inx}
The required sum in the problem is, then, the real part of S. Multiply S with e^ix and subtract from S to obtain
(1-e^{ix})S=e^{ix}+e^{i2x}+e^{i3x}+\ldots + e^{inx} -ne^{i(n+1)x}
\Rightarrow\ (1-e^{ix})S=e^{ix}\dfrac{1-e^{inx}}{1-e^{ix}} -ne^{i(n+1)x}
As such we obtain,
S=e^{ix}\dfrac{1-e^{inx}}{(1-e^{ix})^2} -\dfrac{ne^{i(n+1)x}}{1-e^{ix}}
Taking the real part and simplifying, we get the required sum
\sum_{k=1}^nk\cos kx = \dfrac{n}{2}\ \dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{x}{2}}-\dfrac{1}{2}\ \dfrac{\sin^2\frac{nx}{2}}{\sin^2\frac{x}{2}}

2. Apply Rolle's theorem to the function
f(x)=ax^n+bx^{n-1}+cx
in the interval [0,1].

6. a) The correct inequality should be
(b-a)\sec^2a<\tan b-\tan a < (b-a)\sec^2b
To see this, we see that in (0,\pi/2), sec²x is an increasing function. As such, we get
\sec^2a< \dfrac{1}{(b-a)}\int_a^b \sec^2x\ \mathrm dx< \sec^2b
from where the result follows.

5 >
let f(x)=x+ax^-2
so for maxima or minima , lets put f ' (x) = 0
we get 1 - 2ax^-3 =0
or x =(2a)^1/3
again f '' (x) = 6ax^-4 > 0 for all positive x and all natural a .
so x = (2a)^1/3 is the point of global minima .
thus (2a)^1/3 + a . (2a)^1/3 > 2
so we get a >32 / 27
or a >= 2
as a is a natural number
done

4th

http://www.mathlinks.ro/viewtopic.php?p=3845#3845