1. **Stating the problem:** We are given an expression involving combinations (binomial coefficients) and algebraic terms: $$\text{LHS} = C(k+1, 3(k+1)+2) - C(k+2, 3(k+1) + 1) + (k+1) \cdot 3(k+1) - 1$$ We want to simplify or analyze this expression. 2. **Recall the combination formula:** The binomial coefficient is defined as: $$C(n, r) = \frac{n!}{r!(n-r)!}$$ where $n!$ is the factorial of $n$. 3. **Simplify the indices inside the combinations:** Calculate the terms inside the combinations: - $3(k+1) + 2 = 3k + 3 + 2 = 3k + 5$ - $3(k+1) + 1 = 3k + 3 + 1 = 3k + 4$ So the expression becomes: $$C(k+1, 3k+5) - C(k+2, 3k+4) + (k+1) \cdot 3(k+1) - 1$$ 4. **Analyze the binomial coefficients:** Note that for $C(n, r)$, if $r > n$, then $C(n, r) = 0$ because you cannot choose more elements than available. Here, $3k + 5$ and $3k + 4$ are both greater than $k+1$ and $k+2$ respectively for all $k \geq 0$. Since $3k + 5 > k + 1$ and $3k + 4 > k + 2$, both $C(k+1, 3k+5)$ and $C(k+2, 3k+4)$ are zero. 5. **Simplify the expression using this fact:** $$\text{LHS} = 0 - 0 + (k+1) \cdot 3(k+1) - 1 = 3(k+1)^2 - 1$$ 6. **Final simplified form:** $$\boxed{3(k+1)^2 - 1}$$ This is the simplified expression for the given LHS. --- **Summary:** The binomial coefficients vanish because their lower indices exceed the upper indices, leaving only the algebraic term to simplify.