What is expand (2a+3b)^4 ?

The solution to expand (2a+3b)^4 is 16a^4+96a^3b+216a^2b^2+216ab^3+81b^4

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expand (2a+3b)^4

expand

(2 a + 3 b)^{4}

16 a^{4} + 96 a^{3} b + 216 a^{2} b^{2} + 216 a b^{3} + 81 b^{4}

Solution steps

(2 a + 3 b)^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2 a, b = 3 b

= i = 0 \sum 4 (i 4) (2 a)^{(4 - i)} (3 b)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (2 a)^{4} (3 b)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (2 a)^{3} (3 b)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (2 a)^{2} (3 b)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (2 a)^{1} (3 b)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (2 a)^{0} (3 b)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 a)^{4} (3 b)^{0} : 16 a^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 a)^{3} (3 b)^{1} : 96 a^{3} b

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 a)^{2} (3 b)^{2} : 216 a^{2} b^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 a)^{1} (3 b)^{3} : 216 a b^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 a)^{0} (3 b)^{4} : 81 b^{4}

= 16 a^{4} + 96 a^{3} b + 216 a^{2} b^{2} + 216 a b^{3} + 81 b^{4}

(r^{2})^{- \frac{3}{2}}

\frac{\frac{1}{4}}{1 - \frac{1}{4}}

+ n^{2} - 9 n + 14

\frac{1}{3} \times \frac{1}{6}

7 \times (- 6)

What is expand (2a+3b)^4 ?
The solution to expand (2a+3b)^4 is 16a^4+96a^3b+216a^2b^2+216ab^3+81b^4