Why do I get that $0 = 1$ when trying to prove that $\forall n\in\mathbb{N}, \ 3 + 2\left(n + \sum_{i=1}^{2n-1}(i+1)\right) = (2n+1)^2\,?$

Question

I was playing around with numbers and by looking at some similar observations, I found an interesting pattern. For a natural number $n$, it seemed like the following was always true: $$3 + 2\left(n + \sum_{i=1}^{2n-1}(i+1)\right) = (2n+1)^2.$$

How must we be able to prove this, though?

---

My Attempt:

Assume that the equation is true for all natural $n$, then we can similarly write the same for $n+1$.

$$\begin{align} 3 + 2\left(n+1+\sum_{i=1}^{2(n+1)-1}(i+1)\right) &= \big(2(n+1)+1\big)^2 \\ \Leftrightarrow 3 + 2\left(n+1+\sum_{i=1}^{2n+1}(i+1)\right) &= 3 + 2n + 2\left(1+\sum_{i=1}^{2n+1}(i+1)\right) \\ &= (2n+3)^2 \\ \Leftrightarrow (2n+3)^2 - (2n + 3) &= (2n+3)(2n + 3 - 1) \\ &= 2(2n+3)(n+1) \\ &= 2\left(1+\sum_{i=1}^{2n+1}(i+1)\right) \\ \Leftrightarrow (2n+3)(n+1) &= 1+\sum_{i=1}^{2n+1}(i+1) \\ &= \sum_{i=0}^{2n+1}(i+1) \\ &= \sum_{i=1}^{2n+1}i.\end{align}$$ I know that, $$\sum_{i=1}^ki = \frac{k(k+1)}{2}.\tag{The Triangular Numbers}$$ Therefore, by substituting $k = 2n+1$, we get $$\begin{align} \sum_{i=1}^{2n+1}i &= \frac{(2n+1)(2n+2)}{2} \\ &= \frac{2(n+1)(2n+1)}{2} \\ &= (n+1)(2n+1).\end{align}$$ It follows, then, that $(2n+3)(n+1) = (2n+1)(n+1)$ which means that $2n+3 = 2n+1$ and absurdly we get that $n=n+1$ or $0=1$.

I obviously did something wrong, but I don't know where. Can someone help me out please?

Thank you in advance.

Answer 1

The formula you have found is just simply a way of rewriting $(2n+1)^2$.

$$3+2\left(n+\sum_{i=1}^{2n-1}I+1\right)=3+2\left(n+2n-1+\frac{(2n-1)(2n)}{2}\right)$$ $$=3+2(3n-1+2n^2-n)=4n^2+4n+1=(2n+1)^2$$ Therefore $$3+2\left(n+\sum_{i=1}^{2n-1}I+1\right)\equiv (2n+1)^2$$

Answer 2

Hint

Use that $$\sum_{i=1}^k i= \frac {k(k+1)}{2}$$

Hence $$\sum_{i=1}^{2n-1} i= \frac {2n(2n-1)}{2}$$. Hence your expression turns to $$3+2\left(\frac {2n(2n-1)}{2} +(2n-1)+n\right)$$ $$=3+2n(2n-1)+6n-2$$ $$=4n^2+4n+1$$ $$=(2n+1)^2$$