Why does the exponent function $= 2\cdot 1+2\cdot 2+2\cdot 3+\ldots + 2\cdot(k-1) + k = k^2$?

Question

For $1$ it would be $1$ since $1$ is $k$, for $2$ it would be $4$ because $2(1)+2 = 4$ for $3$ it would be $9$ because $2(1) + 2(2) + 3 = 9$, so why is this? I would also like to add that the series is convergent to $1+3+5+\ldots+(2n-1)$.

Answer 1

Prove by induction. Suppose it is true of $k$. Then

$2(1+2+\cdots+k)+(k+1)=[2(1+2+\cdots+(k-1))+k]+k+(k+1)=k^2+2k+1=(k+1)^2$

Answer 2

I believe what you're asking is why the identity

$$2(1+2+\cdots + (k-1)) + k = k^2$$

holds for all positive integers $k$. This follows directly identity $1+ 2 + \cdots + n = \frac{n(n+1)}{2}$, since

$$ 2(1 + 2 + \cdots + (k-1)) = 2\frac{(k-1)k}{2} = k^2 - k $$

so

$$ 2(1 + 2 + \cdots + (k-1)) + k = (k^2 - k) + k = \boxed{k^2} .$$

Answer 3

Your series can be represented as $\;n+\left(2\left(\sum_\limits{k=1}^{n-1}k\right)\right)$. Using $\sum_\limits{k=1}^{n}k=\frac{n(n+1)}2$, we have $n+2\left(\frac{n(n-1)}2\right)=n+n(n-1)=n+n^2-n=n^2$