What is expand (n+1)^2(2/3)^n ?

The solution to expand (n+1)^2(2/3)^n is (2/3)^nn^2+(2/3)^n*2n+(2/3)^n

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expand (n+1)^2(2/3)^n

expand

(n + 1)^{2} (\frac{2}{3})^{n}

(\frac{2}{3})^{n} n^{2} + (\frac{2}{3})^{n} \cdot 2 n + (\frac{2}{3})^{n}

Solution steps

(n + 1)^{2} (\frac{2}{3})^{n}

= (\frac{2}{3})^{n} (n + 1)^{2}

Expand

(n + 1)^{2} : n^{2} + 2 n + 1

= (\frac{2}{3})^{n} (n^{2} + 2 n + 1)

Apply the distributive law:

m (a + b + c) = ma + mb + m c

(\frac{2}{3})^{n} (n^{2} + 2 n + 1) = (\frac{2}{3})^{n} n^{2} + (\frac{2}{3})^{n} \cdot 2 n + (\frac{2}{3})^{n} \cdot 1

= (\frac{2}{3})^{n} n^{2} + (\frac{2}{3})^{n} \cdot 2 n + (\frac{2}{3})^{n} \cdot 1

Apply rule:

a \cdot 1 = a

(\frac{2}{3})^{n} \cdot 1 = (\frac{2}{3})^{n}

= (\frac{2}{3})^{n} n^{2} + (\frac{2}{3})^{n} \cdot 2 n + (\frac{2}{3})^{n}

9 z^{2} (4 e + 7 f) + 4 e + 7 f

(5 a + b^{5}) l e f t (5 a - b^{5})

\frac{( 2 x + 3 ) ^{3}}{x ^{2} + 4}

(x + 4) (x + \frac{5}{6})

\frac{( x - 4 ) ^{2}}{x + 2}

What is expand (n+1)^2(2/3)^n ?
The solution to expand (n+1)^2(2/3)^n is (2/3)^nn^2+(2/3)^n*2n+(2/3)^n