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Evaluate the sum:

$$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{2k}}\binom{2k}{k}\frac{1}{k+10}$$

Recall the well-known generating function
$$\sum_{k=0}^\infty\,\binom{2k}{k}x^k = \frac{1}{\sqrt{1 -4x}}.$$
Replacing $x$ by $-x^2$ yields
$$\sum_{k=0}^\infty\,(-1)^k\binom{2k}{k}x^{2k} = \frac{1}{\sqrt{1 +4x^2}}.$$
Multiplying $x^{19}$ on each side and then integrating with respect to $x$ from $0$ to $1/2$ gives
$$\sum_{k=0}^\infty\,(-1)^k\frac{\binom{2k}{k}}{(2k+20)2^{2k+20}} = \int_0^{1/2}\,\frac{x^{19}}{\sqrt{1 +4x^2}}\,dx.$$
By the trig. substitution $x = \frac{1}{2}\,\tan\theta$, we find that
$$I:=\int_0^{1/2}\,\frac{x^{19}}{\sqrt{1 +4x^2}}\,dx = \frac{1}{2^{20}}\,\int_0^{\pi/4}\,\tan^{19}\theta\sec\theta\,d\theta.$$
Either use
$$\int_0^{\pi/4}\,\tan^{19}\theta\sec\theta\,d\theta = \int_0^\sqrt{2}\,(u^2-1)^9\,du$$
or by mathematica, we have
$$I = \frac{1}{2^{20}}\,\left(\frac{65536}{230945} - \frac{40427}{230945}\sqrt{2}\right),$$
and so
$$\sum_{k=0}^\infty\,(-1)^k\frac{\binom{2k}{k}}{2^{2k}\,(k+10)} = 2^{21}\cdot I = \frac{1}{230945}\,(131072 -80854\sqrt{2}).$$