Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$

Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.

The $n$th or last term becomes $(4(n+1)-1)=4n+3$.

We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$

The right side is $3(n+1)^2 = 3(n^2 + 2n +1 ) $

Next thing is I appear to be missing the first term in the sum $(2n+1)$ on the left.
Adding $(2n+1)$ to both sides and subtracting $4n+3$ from both sides we get the $n$ case that equals $ 3n^2 $ on the left and

$$3n^2 + 6n +3 + 2n+1 -4n -3= 3n^2 + 4n+1$$

Which leaves me with $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2 +4n +1$

Which is approximately $4n+1$ on the right bigger than what I started with and is exactly the same on the left algebraically I must of done something wrong.

Which leads me to my question, what was it?


$\sum_\limits{k=1}^{n} (2(n+k)-1) = 3n^2$

Base case: $n = 1$

$(2(1+1) - 1) = 3$


$\sum_\limits{k=1}^{n} (2(n+k)-1) = 3n^2$

We will show that $\sum_\limits{k=1}^{n+1} (2((n+1)+k)-1) = 3(n+1)^2$ based on the inductive hypothesis

$\sum_\limits{k=1}^{n+1} (2((n+1)+k)-1)\\ \sum_\limits{k=1}^{n} (2(n+1)+k)-1) + 4n+3\\ \big(\sum_\limits{k=1}^{n} (2(n+k)-1)\big)+\big(\sum_\limits{k=1}^{n}2\big)+ 4n+3\\ 3n^2 + 2n + 4n+3\\ 3(n+1)^2$



The left hand side is the sum of the odd numbers from $2n+1$ up to $4n-1$. Thus when going $n\to n+1$, the first summand $2n+1$ is dropped, and two summands $4n+1$ and $4n+3$ are appended a the end.

  • $\begingroup$ i know what generates the 4n+3 term its form the term n which term generates 4n+1 as the n-th term would be the term 4n-1 $\endgroup$ – Faust Sep 9 '16 at 2:56
  • 4
    $\begingroup$ @Faust7: You should think of it as a range from $2n+1$ to $4n-1$, not that every term gets increased. When you substitute $n \to n+1$ the range becomes $2(n+1)+1$ to $4(n+1)-1$ or $2n+3$ to $4n+3$. That has one more number in it. $\endgroup$ – Ross Millikan Sep 9 '16 at 2:59
  • $\begingroup$ omg you are genius it makes perfect sense now thank you so much! $\endgroup$ – Faust Sep 9 '16 at 3:00

You can try this approach: Calculate sum of first $k$ odd numbers, namely from 1 to $2k-1$.

Denoting it by $S_{k}$. By induction prove this is $k^2$.

Now the desired sum is $S_{2n}-S_n= 4n^2-n^2=3n^2$.


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