What is expand (x+2)^7 ?

The solution to expand (x+2)^7 is x^7+14x^6+84x^5+280x^4+560x^3+672x^2+448x+128

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expand (x+2)^7

expand

(x + 2)^{7}

x^{7} + 14 x^{6} + 84 x^{5} + 280 x^{4} + 560 x^{3} + 672 x^{2} + 448 x + 128

Solution steps

(x + 2)^{7}

Apply binomial theorem:

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

a = x, b = 2

= i = 0 \sum 7 (i 7) x^{(7 - i)} \cdot 2^{i}

Expand summation

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

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} x^{7} \cdot 2^{0} : x^{7}

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} x^{6} \cdot 2^{1} : 14 x^{6}

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} x^{5} \cdot 2^{2} : 84 x^{5}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} x^{4} \cdot 2^{3} : 280 x^{4}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} x^{3} \cdot 2^{4} : 560 x^{3}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} x^{2} \cdot 2^{5} : 672 x^{2}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} x^{1} \cdot 2^{6} : 448 x

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} x^{0} \cdot 2^{7} : 128

= x^{7} + 14 x^{6} + 84 x^{5} + 280 x^{4} + 560 x^{3} + 672 x^{2} + 448 x + 128

x 28 x^{2} - 4 63 x^{4}

6 \times (- 7)

6 \times 10

\frac{1}{3} \times (- 9)

25 x^{2} + 25 x + 6

What is expand (x+2)^7 ?
The solution to expand (x+2)^7 is x^7+14x^6+84x^5+280x^4+560x^3+672x^2+448x+128