Question Appendix.3: Using Mathematical Induction Use mathematical induction to p...

1.  When n = 1, the formula is valid, because

$S_1 = 1^ 2 = \frac{1 (1  +  1)[2(1)  +  1]}{6} = \frac{1(2)(3)}{6} = 1$.

2.  Assuming the formula is true for k,

you must show that it is true for k + 1

$S_{k+1}  =  \frac{(k + 1)[(k + 1) + 1][2(k + 1) + 1]}{6} = \frac{(k + 1)(k + 2)(2k + 3)}{6}$.

To do this, write $S_{k + 1}$ as the sum of $S_k$ and the $(k + 1)$ st term, $(k + 1)^{2}$, as follows.

Combining the results of parts (1) and (2), you can conclude by mathematical induction that the formula is valid for all positive integers n.