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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.

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

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

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  • $\begingroup$ 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

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