Prove: displaystyleΣ_(k=1)^n(n+1-k)cos kx=((n+1)cos…

Prove: $\ \displaystyle\sum_{k=1}^n(n+1-k)\cos kx=\frac{(n+1)\cos x-\cos(n+1)x-n}{4\sin^2\frac x2}$.

Step 1 · Rewrite the triangular weight as a double sum

The weight $n+1-k$ counts how many integers $m$ satisfy $k\le m\le n$, so:

$$ \sum_{k=1}^n(n+1-k)z^k=\sum_{m=1}^n\sum_{k=1}^mz^k=\sum_{m=1}^n\frac{z(1-z^m)}{1-z} =\frac{z}{1-z}\left[n-\sum_{m=1}^nz^m\right] =\frac{z}{1-z}\left[n-\frac{z(1-z^n)}{1-z}\right] $$

Step 2 · Combine over a common denominator

$$ \sum_{k=1}^n(n+1-k)z^k=\frac{nz(1-z)-z^2(1-z^n)}{(1-z)^2}=\frac{nz-(n+1)z^2+z^{n+2}}{(1-z)^2} $$

Step 3 · Substitute $z=e^{ix}$ and factor the denominator

Using $1-z=1-e^{ix}=-2i\sin\dfrac x2\,e^{ix/2}$, so $(1-z)^2=-4\sin^2\dfrac x2\,e^{ix}$. Dividing numerator and denominator by $e^{ix}=z$:

$$ \sum_{k=1}^n(n+1-k)z^k=\frac{e^{i(n+1)x}-(n+1)e^{ix}+n}{-4\sin^2\frac x2} $$

Step 4 · Take the real part

$$ \sum_{k=1}^n(n+1-k)\cos kx=\text{Re}\left(\sum_{k=1}^n(n+1-k)z^k\right) =\frac{\cos(n+1)x-(n+1)\cos x+n}{-4\sin^2\frac x2} $$

$$ \boxed{\sum_{k=1}^n(n+1-k)\cos kx=\frac{(n+1)\cos x-\cos(n+1)x-n}{4\sin^2\frac x2}} $$

Verification, $n=1$

The formula gives $\cos x=\dfrac{2\cos x-\cos2x-1}{4\sin^2\frac x2}$. Since $\cos2x=1-2\sin^2x=1-8\sin^2\frac x2\cos^2\frac x2$ and $\cos x=1-2\sin^2\frac x2$, the numerator becomes $2\cos x-\cos2x-1=4\sin^2\frac x2\cos x$, matching exactly.

The general identity was checked symbolically for $n=1,2,3,4,5$, both for the complex generating-function identity in Step 2 and the final real-part formula.