Sum of Two Binomial Expansions

Observe that

$$3^{10-r}=(\sqrt3)^{20-2r}$$

Therefore

$$S=\sum_{r=0}^{10}\binom{20}{2r}(\sqrt3)^{20-2r}(-1)^r$$

Now use the standard identity

$$\frac{(a+b)^{20}+(a-b)^{20}}{2}=\sum_{r=0}^{10}\binom{20}{2r}a^{20-2r}b^{2r}$$

Take

$$a=\sqrt3,\qquad b=i$$

Since

$$i^{2r}=(-1)^r$$

we get

$$S=\frac{(\sqrt3+i)^{20}+(\sqrt3-i)^{20}}{2}$$

Now write

$$\sqrt3+i=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)$$

Hence by De Moivre's theorem,

$$\left(\sqrt3+i\right)^{20}=2^{20}\left(\cos\frac{20\pi}{6}+i\sin\frac{20\pi}{6}\right)$$

Similarly,

$$\left(\sqrt3-i\right)^{20}=2^{20}\left(\cos\frac{20\pi}{6}-i\sin\frac{20\pi}{6}\right)$$

Adding,

$$S=\frac{2^{20}\left(2\cos\frac{20\pi}{6}\right)}{2}$$

$$=2^{20}\cos\frac{20\pi}{6}$$

$$=2^{20}\cos\frac{10\pi}{3}$$

$$=2^{20}\cos\frac{4\pi}{3}$$

$$=2^{20}\left(-\frac12\right)$$

$$=-2^{19}$$

Final Answer :

$$-2^{19}$$