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Mathematical Induction for the Sum of Squares

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The formula for the sum of square numbers,

$\displaystyle A(n):\ \sum_{i=1}^{n} i^2 = 1^2+2^2+\dots+n^2= \frac{1}{6}n(n+1)(2n+1) \,, $

can be proved by Mathematical Induction.

Base step ($ A(1)$):

$\displaystyle \sum_{i=1}^{1} i^2 = 1^2 = \frac{1\cdot2\cdot3}{6} \,. $

Conclusion ( $ A(n)\implies A(n+1))$:

$\displaystyle \sum_{i=1}^{n+1} i^2$ $\displaystyle =$ $\displaystyle \sum_{i=1}^{n} i^2 + (n+1)^2 = \underbrace{\displaystyle\frac{n(n+1)(2n+1)}{6}}_{ A(n)} + (n+1)^2$  
  $\displaystyle =$ $\displaystyle \displaystyle\frac{(n+1)\big[n(2n+1)+6(n+1)\big]}{6} = \displaystyle\frac{(n+1)(n+2)(2n+3)}{6}\,.$  

As indicated, the induction hypothesis has been applied to obtain the third equality.

(Authors: Kimmerle/Abele)
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  automatisch erstellt am 11.  6. 2007