1. ## Induction help

Use the mathematical induction to show the following. For all integer n=>2

2. Originally Posted by slipz418
Use the mathematical induction to show the following. For all integer n=>2
This exercise occurs so frequently that you can find it practically anywhere.

3. $\displaystyle \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$

If this is so, then

$\displaystyle \sum_{k=1}^{n+1}k^2\ should\ =\frac{(n+1)([n+1]+1)(2[n+1]+1)}{6}$

therefore

P(n)

$\displaystyle \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$

P(n+1)

$\displaystyle \sum_{k=1}^{n+1}k^2=\frac{(n+1)(n+2)(2n+3)}{6}$

Proof

Express P(n+1) in terms of P(n)
so that if P(n) really is true, then P(n+1) must also be true
(P(n) being true causes P(n+1) to be true).

$\displaystyle \sum_{k=1}^{n+1}k^2=\sum_{k=1}^nk^2+(n+1)^2$

which, if P(n) is true will equal

$\displaystyle \frac{n(n+1)(2n+1)}{6}+(n+1)(n+1)=\frac{n(n+1)(2n+ 1)+6(n+1)(n+1)}{6}$

$\displaystyle =\frac{(n+1)[n(2n+1)+6n+6]}{6}=\frac{(n+1)(2n^2+7n+6)}{6}$

$\displaystyle =\frac{(n+1)(2n+3)(n+2)}{6}$

Therefore, if the sum is valid for n=1, it's valid for n=2,
if it's valid for n=2, it's valid for n=3,
if it's valid for n=3, then it's valid for n=4
all the way to infinity,

hence if it's valid for n=1, it's valid for all n.

$\displaystyle \frac{1(1+1)(2+1)}{6}=\frac{6}{6}=1=1^2$

Therefore the equation for the sum of squares is valid