Mathematics Induction

Question

In Mathematical Induction there's something I'm not getting when proving p(k+1) is true Sometimes we just add k+1 to the equation as in

{ P(k)= 1 + 3 + 5 + ... + (2k−1)

P(k+1)= 1 + 3 + 5 + ... + (2k−1) + (2(k+1)−1) }

and { P(k)= 1^3+2^3+...+k^3

P(k+1)=1^3+2^3+...+k^3+(k+1)^3 }

other times we just substitute k+1 in k like here {

P(k)= 1^2 +2^2 +3^2 +(2k)^2

P(k+1)=1^2 +2^2 +3^2 +···+(2(k+1))^2 }

and here

{ p(K)= 1+4+7+···+(3k−2)

p(k+1)= 1+4+7+···+(3(k+1)−2) }

I hope anyone help me understand when and why we substitute K+1 or add K+1

Answer 1

In your first few examples, $P(k)$ is the sum of $k$ terms, so $P(k+1)$ is those $k$ terms plus one more. But with $P(k):=\sum_{j=1}^{2k}j^2$, the number of terms is $2k$, so $P(k+1)$ adds two terms. If you struggle with this kind of thing, ask yourself what $P(k+1)-P(k)$ should be by the definition of $P$. Calculating the result for small $k$ may help you.

Answer 2

You're essentially doing the same in both situations.

Take the following function for example

$$f(k) = 1 + 2 + ... + k$$

When you take $f(k+1)$, you can write that down as

$$f(k+1) = 1 + 2 + ... + (k+1)$$

Notice that this is the same as

$$f(k+1) = 1 + 2 + ... + k + (k+1)$$

Because the ellipse sign (...) expands to a sum including k.

Answer 3

It seems you're just getting confused about what is hidden by the $\cdots$, try writing out one or two more terms after the dots, or better (but possibly harder) find and use an exact representation of the ecpressions.