What is expand (2e+f)^7 ?

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

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expand (2e+f)^7

expand

(2 e + f)^{7}

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

Solution steps

(2 e + f)^{7}

Apply binomial theorem:

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

a = 2 e, b = f

= i = 0 \sum 7 (i 7) (2 e)^{(7 - i)} f^{i}

Expand summation

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

\frac{7 !}{0 ! ( 7 - 0 ) !} (2 e)^{7} f^{0} = 128 e^{7}

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} (2 e)^{6} f^{1} : 448 e^{6} f

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} (2 e)^{5} f^{2} : 672 e^{5} f^{2}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} (2 e)^{4} f^{3} : 560 e^{4} f^{3}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} (2 e)^{3} f^{4} : 280 e^{3} f^{4}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} (2 e)^{2} f^{5} : 84 e^{2} f^{5}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} (2 e)^{1} f^{6} : 14 e f^{6}

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} (2 e)^{0} f^{7} : f^{7}

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

Rewrite in standard form

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

3 (2 i, - 4 j) + 2 (- 5 i, - 2 j)

- 1^{2} - 2

\frac{- x + 4}{6 ( x + 2 )}

\frac{d ^{2}}{d x ^{2}} ((y) (3 x - 4))

3 x (2 e^{x - 5 y} y - 5 e^{x - 5 y} y^{2})

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