The base cases n = 1, 2, 3 can be verified to be true. Assume it is true for some n = k. We want to show that
$\displaystyle (k+1) + 2k + 3(k-1) + ... + (k-1)3 + k(2) + (k+1) = \frac{(k+1)(k+2)(k+3)}{6}$
Using our useful "hint," rewrite the LHS as
$\displaystyle k + 2(k-1) + 3(k-2) + ... + (k-1)2 + k + (1+2+...+k+(k+1))$
$\displaystyle =\frac{k(k+1)(k+2)}{6} + (1+2+...+k+(k+1))$
$\displaystyle =\frac{k(k+1)(k+2)}{6} + \frac{(k+1)(k+2)}{2}$
$\displaystyle =\frac{k(k+1)(k+2) + 3(k+1)(k+2)}{6}$
$\displaystyle =\frac{(k+1)(k+2)(k+3)}{6}$
The sum of the integers from $\displaystyle 1$ to $\displaystyle a$ is $\displaystyle \frac{a(a+1)}{2}$ (which can also be proved by induction!)
Therefore the sum of the integers from $\displaystyle 1$ to $\displaystyle k+1$ is $\displaystyle \frac{(k+1)(k+2)}{2}$.