en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (i+1)^7

Solution

simplify

(i + 1)^{7}

Solution

8 - 8 i

Solution steps

(i + 1)^{7}

Apply binomial theorem:

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

a = i, b = 1

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

Simplify

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

= - 8 i + 8

Rewrite in standard complex form:

8 - 8 i

= 8 - 8 i

Popular Examples

expand (dx)/((7x+3)^3)expand

\frac{d x}{( 7 x + 3 ) ^{3}}

simplify (x-10)(x+sqrt(5))(x-sqrt(5))simplify

(x - 10) (x + 5) (x - 5)

expand 1/((s-1)^3)expand

\frac{1}{( s - 1 ) ^{3}}

simplify (x-2)(x-1+i)(x+1-i)simplify

(x - 2) (x - 1 + i) (x + 1 - i)

expand 1/((1+jw)(3+jw))expand

\frac{1}{( 1 + j w ) ( 3 + j w )}

Frequently Asked Questions (FAQ)

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