Animated Solution for Mathematics - Binomial Theorem: The value of ${(}^{21}{C}_1{-}^{10}{C}_1)+{(}^{21}{C}_2{-}^{10}{C}_2)+{(}^{21}{C}_3{-}^{10}{C}_3)+{(}^{21}{C}_4{-}^{10}{C}_4)+\cdots +{(}^{21}{C}_{10}{-}^{10}{C}_{10})$ is:

The problem asks us to evaluate the expression:

At first glance, this appears to be a daunting list of terms. However, we can apply the power of mathematical decomposition to simplify the structure.

The beauty of the summation operator $\sum$ lies in its linearity. We can treat the difference of two terms as the difference of two separate sums:

Now, instead of one complex expression, we have two manageable components. Let us define:

Let us focus on $S_1$. We know the fundamental identity:

For $n=21$, the total sum is $2^{21}$. Because $21$ is an odd number, the binomial coefficients are perfectly symmetric, satisfying ${}^{21}{C}_k{=}^{21}{C}_{21-k}$.

This symmetry implies that the sum of the first half (from $k=0$ to $10$) is exactly equal to the sum of the second half (from $k=11$ to $21$). Since the total sum is $2^{21}$, each half must be:

Thus, \sum_{k=0}^{10} ^{21}C_k = 2^{20}. Since our sum $S_1$ starts at $k=1$, we must subtract the $k=0$ term, which is ${}^{21}{C}_0=1$:

Now, we tackle S_2 = \sum_{k=1}^{10} ^{10}C_k. The total sum \sum_{k=0}^{10} ^{10}C_k is $2^{10}$.

Again, our sum starts at $k=1$, so we must subtract the $k=0$ term, ${}^{10}{C}_0=1$:

Finally, we bring it all together by substituting our values back into $S=S_1-S_2$:

As we expand the brackets, the $-1$ and the $+1$ cancel out perfectly. We are left with the elegant final result: