Let's say we are denoting the sum of $n$ odd numbers. Then in symbols $1+3+5+\cdots+(2n-1)$.

If we substitute $(k+1)$ for $n$. $2n-1=2k+1$ So $1+3+5+...+2k+1$

Then can we use $1+3+5+\cdots+(2n+1)$ instead of $1+3+5+\cdots+(2n-1)$?

Logically, I think they are the same, but when I think of the number of terms, $1+3+5+\cdots+(2n+1)$ has one more term than $1+3+5+\cdots+(2n-1)$.

I also think $1+3+5+\cdots+(2n+1)$ is equal to $1+3+5+\cdots+(2n-1)+(2n+1)$, so it would be contradiction. But I don't know how to explain $1+3+5+\cdots+(2n-1)=1+3+5+\cdots+2k+1$ is a contradiction.


2 Answers 2


If you're writing $1+3+5+\cdots+45$, for instance, you could describe that as

$1 + 3 + 5 + \cdots + (2n-1)$ for $1 \le n \le 23$, and there are 23 terms.

You could also describe it as

$1 + 3 + 5 + \cdots + (2n+1)$, but this time we have $0 \le n \le 22$, and there are still only 23 terms: even though the last value of $n$ is 22, this time we're also counting $n=0$.


Suppose your inductive hypothesis is $$1+3+5+...+(2n-1) = n^2.$$

You initially show the inductive hypothesis is true when $n=1$, e.g. by saying $1=1^2$ is indeed correct

Then you assume the inductive hypothesis is true for $n=k$ so $1+3+5+...+(2k-1) = k^2.$ You add $(2k+1)$ to both sides so the left hand side becomes $1+3+5+...+(2k+1)$ and the right hand side becomes $k^2 +2k +1 = (k+1)^2$. Since these are equal, this shows the inductive hypothesis would then be true for $n=k+1$.

So using the axiom of induction, this proves the inductive hypothesis.

There is no need for contradiction.


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