# Well, just apply induction

**Algebra**Level 2

\[\large \displaystyle\sum_{r=1}^n r = \dfrac{n(n+1)}{2}\]

It is well known that the formula for the sum of the first \(n\) natural numbers is given as above. Keeping in mind of the formula stated above, is the summation below true?

\[\large\displaystyle\sum_{r=1}^n (r+1) = \dfrac{(n+1)(n+2)}{2}\]