What is expand (1+n)^9 ?

The solution to expand (1+n)^9 is 1+9n+36n^2+84n^3+126n^4+126n^5+84n^6+36n^7+9n^8+n^9

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expand (1+n)^9

expand

(1 + n)^{9}

1 + 9 n + 36 n^{2} + 84 n^{3} + 126 n^{4} + 126 n^{5} + 84 n^{6} + 36 n^{7} + 9 n^{8} + n^{9}

Solution steps

(1 + n)^{9}

Apply binomial theorem:

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

a = 1, b = n

= i = 0 \sum 9 (i 9) \cdot 1^{(9 - i)} n^{i}

Expand summation

= \frac{9 !}{0 ! ( 9 - 0 ) !} \cdot 1^{9} n^{0} + \frac{9 !}{1 ! ( 9 - 1 ) !} \cdot 1^{8} n^{1} + \frac{9 !}{2 ! ( 9 - 2 ) !} \cdot 1^{7} n^{2} + \frac{9 !}{3 ! ( 9 - 3 ) !} \cdot 1^{6} n^{3} + \frac{9 !}{4 ! ( 9 - 4 ) !} \cdot 1^{5} n^{4} + \frac{9 !}{5 ! ( 9 - 5 ) !} \cdot 1^{4} n^{5} + \frac{9 !}{6 ! ( 9 - 6 ) !} \cdot 1^{3} n^{6} + \frac{9 !}{7 ! ( 9 - 7 ) !} \cdot 1^{2} n^{7} + \frac{9 !}{8 ! ( 9 - 8 ) !} \cdot 1^{1} n^{8} + \frac{9 !}{9 ! ( 9 - 9 ) !} \cdot 1^{0} n^{9}

Simplify

\frac{9 !}{0 ! ( 9 - 0 ) !} \cdot 1^{9} n^{0} : 1

Simplify

\frac{9 !}{1 ! ( 9 - 1 ) !} \cdot 1^{8} n^{1} : 9 n

Simplify

\frac{9 !}{2 ! ( 9 - 2 ) !} \cdot 1^{7} n^{2} : 36 n^{2}

Simplify

\frac{9 !}{3 ! ( 9 - 3 ) !} \cdot 1^{6} n^{3} : 84 n^{3}

Simplify

\frac{9 !}{4 ! ( 9 - 4 ) !} \cdot 1^{5} n^{4} : 126 n^{4}

Simplify

\frac{9 !}{5 ! ( 9 - 5 ) !} \cdot 1^{4} n^{5} : 126 n^{5}

Simplify

\frac{9 !}{6 ! ( 9 - 6 ) !} \cdot 1^{3} n^{6} : 84 n^{6}

Simplify

\frac{9 !}{7 ! ( 9 - 7 ) !} \cdot 1^{2} n^{7} : 36 n^{7}

Simplify

\frac{9 !}{8 ! ( 9 - 8 ) !} \cdot 1^{1} n^{8} : 9 n^{8}

Simplify

\frac{9 !}{9 ! ( 9 - 9 ) !} \cdot 1^{0} n^{9} : n^{9}

= 1 + 9 n + 36 n^{2} + 84 n^{3} + 126 n^{4} + 126 n^{5} + 84 n^{6} + 36 n^{7} + 9 n^{8} + n^{9}

2 (\frac{5 x - ( 1 )}{( x )})^{2} - 3 (\frac{5 x - ( 1 )}{( x )}) - 2

\frac{( 4 0 ^{55} ) ^{'} 5 3 ^{^{''}} n}{( 7 1 ^{53} ) ^{'} 1 ^{^{''}} e}

- 19 x + 5 - 6 x - 20

\frac{( 30 - 20 x + 11 x ^{2} + 19 x ^{3} + 4 x ^{4} )}{( x + 3 )}

(a + b)^{3} - 8 b^{3}

What is expand (1+n)^9 ?
The solution to expand (1+n)^9 is 1+9n+36n^2+84n^3+126n^4+126n^5+84n^6+36n^7+9n^8+n^9