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Popular >

expand (1+i)^7

Solution

expand

(1 + i)^{7}

Solution

8 - 8 i

Solution steps

(1 + i)^{7}

Apply binomial theorem:

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

a = 1, b = i

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

Simplify

1 + 7 i + 21 i^{2} + 35 i^{3} + 35 i^{4} + 21 i^{5} + 7 i^{6} + i^{7} : - 8 i + 8

= - 8 i + 8

Rewrite in standard complex form:

8 - 8 i

= 8 - 8 i

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Frequently Asked Questions (FAQ)

What is expand (1+i)^7 ?
The solution to expand (1+i)^7 is 8-8i