Mathematics - Binomial Theorem Question with Solution | TestHub

We are given the sum $S_n={}^{3n}{C}_0-{}^{3n}{C}_1+{}^{3n}{C}_2+\dots +(-1{)}^n\cdot {}^{3n}{C}_n$.

Using the identity ${}^n{C}_r={}^{n-1}{C}_{r-1}+{}^{n-1}{C}_r$, we can rewrite each term:

$${}^{3n}{C}_0={}^{3n-1}{C}_0$$

$$-{}^{3n}{C}_1=-({}^{3n-1}{C}_0+{}^{3n-1}{C}_1)=-{}^{3n-1}{C}_0-{}^{3n-1}{C}_1$$

$${}^{3n}{C}_2={}^{3n-1}{C}_1+{}^{3n-1}{C}_2$$

$$-{}^{3n}{C}_3=-({}^{3n-1}{C}_2+{}^{3n-1}{C}_3)=-{}^{3n-1}{C}_2-{}^{3n-1}{C}_3$$

This pattern continues until the last term:

$$(-1{)}^n\cdot {}^{3n}{C}_n=(-1{)}^n\cdot ({}^{3n-1}{C}_{n-1}+{}^{3n-1}{C}_n)$$

$$(-1{)}^n\cdot {}^{3n}{C}_n=(-1{)}^n\cdot {}^{3n-1}{C}_{n-1}+(-1{)}^n\cdot {}^{3n-1}{C}_n$$

When we add all these terms, most of them cancel out. For example, the $-{}^{3n-1}{C}_0$ from the second term cancels with the ${}^{3n-1}{C}_0$ from the first term. Similarly, $-{}^{3n-1}{C}_1$ from the second term cancels with ${}^{3n-1}{C}_1$ from the third term, and so on.

The sum $S_n$ simplifies to:

$$S_n=(-1{)}^n\cdot {}^{3n-1}{C}_n$$