What is expand (1+2x)^7 ?

The solution to expand (1+2x)^7 is 1+14x+84x^2+280x^3+560x^4+672x^5+448x^6+128x^7

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expand (1+2x)^7

expand

(1 + 2 x)^{7}

1 + 14 x + 84 x^{2} + 280 x^{3} + 560 x^{4} + 672 x^{5} + 448 x^{6} + 128 x^{7}

Solution steps

(1 + 2 x)^{7}

Apply binomial theorem:

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

a = 1, b = 2 x

= i = 0 \sum 7 (i 7) \cdot 1^{(7 - i)} (2 x)^{i}

Expand summation

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

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 1^{7} (2 x)^{0} : 1

Simplify

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

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 1^{5} (2 x)^{2} : 84 x^{2}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 1^{4} (2 x)^{3} : 280 x^{3}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 1^{3} (2 x)^{4} : 560 x^{4}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 1^{2} (2 x)^{5} : 672 x^{5}

Simplify

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

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 1^{0} (2 x)^{7} : 128 x^{7}

= 1 + 14 x + 84 x^{2} + 280 x^{3} + 560 x^{4} + 672 x^{5} + 448 x^{6} + 128 x^{7}

1128 + 7 + 27

\frac{2}{3} \times 24

\frac{3}{27}

\frac{2}{3} + \frac{4}{4} + \frac{6}{2}

8 X 4^{3}

What is expand (1+2x)^7 ?
The solution to expand (1+2x)^7 is 1+14x+84x^2+280x^3+560x^4+672x^5+448x^6+128x^7