If $$n \geq 2$$ is a positive integer, then the sum of the series $${}^{n+1}C_2 + 2({}^2C_2 + {}^3C_2 + {}^4C_2 + \ldots + {}^nC_2)$$ is

We need to find the sum $${}^{n+1}C_2 + 2\left({}^2C_2 + {}^3C_2 + {}^4C_2 + \ldots + {}^nC_2\right)$$.

First, we compute $${}^{n+1}C_2 = \frac{n(n+1)}{2}$$.

For the inner sum, we apply the hockey stick identity: $${}^2C_2 + {}^3C_2 + \ldots + {}^nC_2 = {}^{n+1}C_3 = \frac{(n+1)n(n-1)}{6}$$.

Substituting into the original expression, we get $$\frac{n(n+1)}{2} + 2 \cdot \frac{(n+1)n(n-1)}{6} = \frac{n(n+1)}{2} + \frac{n(n+1)(n-1)}{3}$$.

Taking $$n(n+1)$$ as a common factor, this becomes $$n(n+1)\left(\frac{1}{2} + \frac{n-1}{3}\right) = n(n+1) \cdot \frac{3 + 2(n-1)}{6} = \frac{n(n+1)(2n+1)}{6}$$.

Therefore, the sum equals $$\frac{n(n+1)(2n+1)}{6}$$.