Prove by induction: $\sum_{k=n+1}^{2n}(2k - 1) = 3n^2$

Question

I found this question on a math textbook:

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

I have to prove this statement with induction.

I want to put it out there as none has posted anything about this exercise, it may be interesting because we can work with both bounds of a summation range and induction.

So how do you prove it?

Answer 1

Anyway, I sat down and got to a correct answer that helped me understand exactly what was going on and why it was going on. I wish to share it with anyone that might be looking for this type of explanation in the future (I ask forgiveness for "bro math"):

$$P(n)=\sum_{k=n+1}^{2n}(2k - 1) = 3n^2, n \in \mathbb{N}$$

This is a summation, meaning you have to sum from $k$ to $n$, in this case from

$$k = n + 1$$

to

$$2n$$

since

$$n \geq 0$$ $$n + 1 \leq 2n$$ $$k \leq 2n$$

This means we are summing any value between these two boundaries:

$$\sum_{k=n+1}^{2n}(2k - 1) = range\{[2(n + 1) - 1] + [2((n + 1) + 1) - 1] + ... + [2(2n - 1) - 1] + [2(2n) - 1]\}$$

To demonstrate it by induction we have to consider the next $n$th value of the sum and see if it respect the $n$th value of the equality, thus if

$$P(n + 1) = \sum_{k=(n+1)+1}^{2(n+1)}(2k - 1) = 3(n+1)^2, n \in \mathbb{N}$$

Here's how it can help think about this problem, we should try to get the sum $\sum_{k=(n+1)+1}^{2(n+1)}(2k - 1)$ to look like the sum of our proposition $\sum_{k=n+1}^{2n}(2k - 1)$, because we can substitute it, by induction hypothesis, with $3n^2$.

Now let's expand and consider what is this sum like at both boundaries:

$$\sum_{k=(n + 1)+1}^{2(n + 1)}(2k - 1) = \sum_{k=n + 2}^{2n + 2}(2k - 1) = range\{[2(n + 2) - 1] + [2((n + 2) + 1) - 1] + ... + [2((2n + 2) - 1) - 1] + [2(2n + 2) - 1]\}$$

Now to get the summation from here: $\sum_{k=n + 2}^{2n + 2}(2k - 1)$ to here $\sum_{k=n + 1}^{2n}(2k - 1)$ we need to do two operations at its boundaries, which are:

1. Decreasing the upper boundary of the sum by two times to go from $2n + 2$ (by $-1$) to $2n + 1$ to $2n$, but by doing this to maintain the "equilibrium" of the sum we need to add any value we eliminate from it (viceversa if we had to increase the upper boundary we would have had to subtract any value we added to the sum):

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

If we represent the summation as a "range" we get:

$$\sum_{k=n + 2}^{2n}(2k - 1) + (2(2n + 1) - 1) + (2(2n + 2) - 1) = range\{[2(n + 2) - 1] + [2((n + 2) + 1) - 1] + ... + [2((2n) - 1) - 1]\} + (2(2n + 1) - 1) + (2(2n + 2) - 1) = range\{[2(n + 2) - 1] + [2((n + 2) + 1) - 1] + ... + [2((2n + 2) - 1) - 1] + [2(2n + 2) - 1]\} = \sum_{k=(n + 1)+1}^{2n + 2}(2k - 1)$$

2. Now the tricky part: We have to Decrease the lower bound of the sum once: We need to go from $\sum_{k=n + 2}^{2n}(2k - 1)$ to $\sum_{k=n + 1}^{2n}(2k - 1)$. This is achieved by subtracting one from the lower bound of the "range": $k = n + 2, k = n + 2 - 1 = n + 1$, thus adding one element to its "range" since: $k = n + 1 < k = n + 2$ and $k = (n + 1) + 1 \Longleftrightarrow k = n + 2$, visually this is the operation we do to the "range":

from $$[2(n + 1) - 1], \sum_{k=n + 2}^{2n}(2k - 1) = [2(n + 1) - 1] + range\{[2(n + 2) - 1] + [2((n + 2) + 1) - 1] + ... + [2((2n) - 1) - 1]\}$$

to $$\sum_{k=n + 1}^{2n}(2k - 1) = range\{[2(n + 1) - 1] + [2(n + 2) - 1] + [2((n + 2) + 1) - 1] + ... + [2((2n) - 1) - 1]\}$$

The same way we equilibrated the sum by adding elements that we eliminated from the upper bound of the range, here, we need to subtract any elements that we just add to the lower bound of the range:

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

As a "range":

$$[2(n + 1) - 1], \sum_{k=n + 2}^{2n}(2k - 1) = [2(n + 1) - 1],\ range\{[2(n + 2) - 1] + [2((n + 2) + 1) - 1] + ... + [2((2n) - 1) - 1]\} = range\{[2(n + 1) - 1] + [2(n + 2) - 1] + [2((n + 2) + 1) - 1] + ... + [2((2n) - 1) - 1]\} - (2(n + 1) - 1)$$

Finally, we have:

$$\sum_{k=(n + 1)+1}^{2(n + 1)}(2k - 1) \Longleftrightarrow$$ $$ \sum_{k=n + 2}^{2n + 1}(2k - 1) + (2(2n + 2) - 1) \Longleftrightarrow$$ $$ \sum_{k=n + 2}^{2n}(2k - 1) + (2(2n + 1) - 1) + (2(2n + 2) - 1) \Longleftrightarrow$$ $$ \sum_{k=n + 1}^{2n}(2k - 1) - (2(n + 1) - 1) + (2(2n + 1) - 1) + (2(2n + 2) - 1) \Longleftrightarrow$$ $$ 3n^2 - (2(n + 1) - 1) + (2(2n + 1) - 1) + (2(2n + 2) - 1) \Longleftrightarrow$$ $$ 3n^2 - (2(n + 1) - 1) + (2(2n + 1) - 1) + (2(2n + 2) - 1) \Longleftrightarrow$$ $$ 3n^2 - (2n + 1) + (4n + 1) + (4n + 3) \Longleftrightarrow$$ $$ 3n^2 - (2n + 1) + (4n + 1) + (4n + 3) \Longleftrightarrow$$ $$ 3n^2 - 2n + 1 + 4n + 1 + 4n + 3 \Longleftrightarrow$$ $$ 3n^2 + 6n + 3 \Longleftrightarrow$$ $$ 3(n^2 + 2n + 1) \Longleftrightarrow$$ $$ 3(n + 1)(n + 1) \Longleftrightarrow$$ $$ 3(n + 1)^2 \Longleftrightarrow$$

I wish that anyone looking for this type of answer will find it, and I hope they'll understand it, and maybe they'll like math a little bit more.

Any error on language / definitions pleas edit the post, thank you.

Answer 2

Base case at $n=1\Rightarrow \sum_{k=2}^2 (2k-1)=3$

Let $m\in\mathbb{N}$ be a number that satisfy the statement ($\sum_{k=m+1}^{2m}(2k-1)=3m^2$).

So, we will see the statement at $m+1$:

$$\sum_{k=m+2}^{2m+2}(2k-1)=\left[2(2m+2)-1\right]+\left[2(2m+1)-1\right]-\left[2(m+1)-1\right]+\sum_{k=m+1}^{2m}(2k-1)=\left[4m+4-1\right]+\left[4m+2-1\right]-\left[2m+2-1\right]+3m^2=3m^2+6m+3=3(m+1)^2 \space \blacksquare$$

So we are done here.