Prove by induction $(n+1) + (n + 2) +\dots + (2n - 1 ) + (2n) = \frac{n(3n + 1)}{2}$, $n \geq 1$

Question

Prove by induction that $(n+1) + (n + 2) +\dots + (2n - 1 ) + (2n) = \dfrac{n(3n + 1)}{2}$ for $n \geq 1$.

I tried to add $(2(k+1)-1) + (2(k+1))$ to both sides and got stuck.

Any other ideas?

Answer 1

Yes, in the inductive step we start from $$(n+1) + (n + 2) +\dots+ (2n – 1 ) + (2n) = n(3n + 1)/2.$$ We add $ (2(n+1)-1) + (2(n+1))$ to both sides AND we subtract $n+1$ from both sides. Then we verify that the right hand side is equal to $$ (n+1)(3(n+1) + 1)/2.$$ Can you take it form here?

Answer 2

If you want to be sneaky, show the easier $s(n) := \sum\limits_{k=1}^n k = \frac{n(n+1)}{2}$ and then subtract $s(n)$ from $s(2n)$.