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**Archie Meade** Part b) attempts to show that __if__ the statement is true for some n=k,

then being true for n=k __causes__ the statement to be true for the next n,

which is n=k+1.

The purpose of this is to show that...

true for n=1 causes the equality to be true for n=2,

true for n=2 causes the equality to be true for n=3,

true for n=3 causes the equality to be true for n=4....

all the way to infinity

then you can see by thinking about it that an infinite chain of cause and effect

has been set up between adjacent terms.

Hence we express the k+1 version in terms of the k version

to see if there is a cause and effect relationship.

Essentially, this is why part b) is written as it is.

$\displaystyle 1+7+19+...+(3k^2-3k+1)=k^3$ ?

If it is then examining

$\displaystyle 1+7+19+...(3k^2-3k+1)+\left[3(k+1)^2-3(k+1)+1\right]$

which is the sum of (k+1) terms,

and examining if this is $\displaystyle (k+1)^3$ __if__ the sum of k terms is $\displaystyle k^3,$ we get

$\displaystyle k^3+3(k+1)^2-3(k+1)+1=k^3+3(k^2+2k+1)-3(k+1)+1$

$\displaystyle =k^3+3k^2+6k+3-3k-3+1=k^3+3k^2+3k+1$

which is $\displaystyle (k+1)^3,$ since

$\displaystyle (k+1)^3=(k+1)^2(k+1)=\left(k^2+2k+1\right)(k+1)$$\displaystyle =k^3+k^2+2k^2+2k+k+1=k^3+3k^2+3k+1$

Hence, the equality being true for some term n=k causes the equality to be true for the next term n=k+1.

Therefore there is an infinite chain of cause and effect.

If the equality is true for n=1, this causes the equality to be true for all n.