Prove by induction, explaining each step carefully, that the sum of the first $2n$ odd positive integers is equal to $4n^2$.

Let P(n) be the statement $P(n)=\sum_{n=1}^{2n} 2n-1 = 4n^2$

The $P(1)$ asserts that $(2(1)-1)+(2(2)-1)=1+3=4$, and we see that P(1) is T, so this establishes the basis for induction.

To verify the induction step, we suppose the P(k) is T, where $k\in \!\,\mathbb{N} \!\,$. That is, we assume: $\sum_{k=1}^{2k} 2k-1 = 4k^2$

(now this is where I am screwing up)

Since we wish to conclude that P(k+1) is T, we add $2k+1$ to both sides.

$\sum_{k=1}^{2k} (2k-1) + (2k+1) = 4k^2 + (2k+1)$ $\sum_{k=1}^{2k} (2k-1) + (2k+1) = 4k^2 + (2k+1)$

  • 3
    $\begingroup$ The $\;n-$th **odd** natural number is $\;(2n-1)\;$ , not $\;2n\;$ , which is even, and the $\;2n-$ th odd natural number is $\;2(2n)-1\;$ $\endgroup$ – Timbuc Nov 3 '14 at 22:05
  • 1
    $\begingroup$ The OP did not mention primes, but probably the last term should be $2n-1$ $\endgroup$ – Peter Nov 3 '14 at 22:06
  • $\begingroup$ It would help if the claim would be mentioned. I am not sure what has to be proven. $\endgroup$ – Peter Nov 3 '14 at 22:08
  • $\begingroup$ If the sum $1+3+5+...+(2n-1)$ has to be calculated, the correct answer would be $n^2$ $\endgroup$ – Peter Nov 3 '14 at 22:11
  • $\begingroup$ It's easier to find and prove a formula for the sum of the first $k$ odd positive integers, show that this sum is equal to $k^2$, and then put in $k=2n$ to show that the sum of the first $2n$ odd positive integers is $(2n)^2$ as desired. $\endgroup$ – MJD Nov 3 '14 at 22:13

You wrote $P(n)=1+3+\cdots +2n$. This is not $P(n)$. For one, that last term is an even number, and you want to only sum odd numbers.

What you want is the first $2n$ odd numbers. The first odd number is $1=2(\color{red}{1})-1$. The second odd number is $3=2(\color{red}{2})-1$. The third odd number is $5=2(\color{red}{3})-1$. The fourth odd number is $7=2(\color{red}{4})-1$. I hope you see the pattern.

So $$P(n)=\sum_{k=1}^{2n} (2k-1) = 1+3+\cdots +4n-1$$

  • $\begingroup$ Would you mind taking a look at my attempt again? I am still struggling :( $\endgroup$ – Math Major Nov 3 '14 at 22:28
  • $\begingroup$ @MathMajor First, be careful about the limits in that sum you wrote: it's $2n$ not $2k$ in $P(n)$. Next, when you write $P(n+1)$ you need to replace all occurrences of $n$ in $P(n)$ by $n+1$. There is only one in this case so you get $P(n+1)=\sum_{k=1}^{2(n+1)}(2k-1) = \sum_{k=1}^{2n}(2k-1)+[2(2(n+1))-1].$ $\endgroup$ – Casteels Nov 4 '14 at 5:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.