n $\sqrt{n}$ + (n+2)$\sqrt{n+1}$

asked by guest
on Nov 25, 2024 at 6:32 am



You asked:

Evaluate the expression: \(n \cdot \sqrt{n} + \left(n + 2\right) \cdot \sqrt{n + 1}\)

MathBot Answer:

Evaluated



\(\displaystyle n \cdot \sqrt{n} + \left(n + 2\right) \cdot \sqrt{n + 1} = n^{\frac{3}{2}} + \sqrt{n + 1} \left(n + 2\right) \)


Expanded

\[n \cdot \sqrt{n} + \left(n + 2\right) \cdot \sqrt{n + 1} = n^{\frac{3}{2}} + n \sqrt{n + 1} + 2 \sqrt{n + 1}\]


Factored

\[n \cdot \sqrt{n} + \left(n + 2\right) \cdot \sqrt{n + 1} = n^{\frac{3}{2}} + n \sqrt{n + 1} + 2 \sqrt{n + 1}\]