Problem on Limits

$\begin{aligned} L & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^{n-1} \cos^{2p} \left(\frac {k\pi}{2n} \right) & \small \blue{\text{By Riemann sums}} \\ & = \int_0^1 \cos^{2p} \left(\frac {\pi x}2 \right) dx & \small \blue{\text{Putting }p=5} \\ & = \int_0^1 \cos^{10} \left(\frac {\pi x}2 \right) dx & \small \blue{\text{Let }u = \frac {\pi x}2 \implies du = \frac \pi 2\ d x} \\ & = \frac 2\pi \int_0^\frac \pi 2 \cos^{10} u \ du & \small \blue{\text{Using reduction formula}} \\ & = \frac 9{5\pi} \int_0^\frac \pi 2 \cos^8 u \ du \\ & = \frac {63}{40\pi} \int_0^\frac \pi 2 \cos^6 u \ du \\ & = \frac {63}{48\pi} \int_0^\frac \pi 2 \cos^4 u \ du \\ & = \frac {63}{64\pi} \int_0^\frac \pi 2 \cos^2 u \ du \\ & = \frac {63}{128\pi} \int_0^\frac \pi 2 (1+\cos (2u)) \ du \\ & = \frac {63}{128\pi} \left[u+\frac {\sin (2u)}2 \right]_0^\frac \p1 2 \\ & = \frac {63}{256} = \frac {7\cdot 3^2}{2^8} \end{aligned}$

Therefore $a+b+c = 7+3+2 = \boxed{12}$ .

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References:

• Riemann sums
• Reduction formula: $\displaystyle \int \cos^m x \ dx = \frac {\sin x \cos^{m-1} x}m + \frac {m-1}m \int \cos^{m-2} x \ dx$

The given sum is equal to $\displaystyle \int_0^1 \cos^{10} \left (\frac{πx}{2}\right ) dx=\dfrac{63}{256}=\dfrac{7\times 3^2}{2^8}$ .

So $a=7,b=3,c=2$ and $a+b+c=\boxed {12}$ .