What is expand (1+x)^7 ?

The solution to expand (1+x)^7 is 1+7x+21x^2+35x^3+35x^4+21x^5+7x^6+x^7

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expand (1+x)^7

expand

(1 + x)^{7}

1 + 7 x + 21 x^{2} + 35 x^{3} + 35 x^{4} + 21 x^{5} + 7 x^{6} + x^{7}

Solution steps

(1 + x)^{7}

Apply binomial theorem:

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

a = 1, b = x

= i = 0 \sum 7 (i 7) \cdot 1^{(7 - i)} x^{i}

Expand summation

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

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 1^{7} x^{0} : 1

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 1^{6} x^{1} : 7 x

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 1^{5} x^{2} : 21 x^{2}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 1^{4} x^{3} : 35 x^{3}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 1^{3} x^{4} : 35 x^{4}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 1^{2} x^{5} : 21 x^{5}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 1^{1} x^{6} : 7 x^{6}

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 1^{0} x^{7} : x^{7}

= 1 + 7 x + 21 x^{2} + 35 x^{3} + 35 x^{4} + 21 x^{5} + 7 x^{6} + x^{7}

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

- \frac{12}{15}

- 8 \cdot 8

1111 + 211

(n - 2)^{3}

What is expand (1+x)^7 ?
The solution to expand (1+x)^7 is 1+7x+21x^2+35x^3+35x^4+21x^5+7x^6+x^7