What is taylor (1+x)^n ?

The answer to taylor (1+x)^n is \ldots

Popular >

taylor (1+x)^n

taylor

(1 + x)^{n}

\dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( ( 1 + x ) ^{n} ) ( - 1 )}{1 !} (x + 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 + x ) ^{n} ) ( - 1 )}{2 !} (x + 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 + x ) ^{n} ) ( - 1 )}{3 !} (x + 1)^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} (x + 1) + \frac{0}{2 !} (x + 1)^{2} + \frac{0}{3 !} (x + 1)^{3} + \frac{0}{4 !} (x + 1)^{4} + \frac{0}{5 !} (x + 1)^{5} + \frac{0}{6 !} (x + 1)^{6} + \frac{0}{7 !} (x + 1)^{7} + \frac{0}{8 !} (x + 1)^{8} + \frac{0}{9 !} (x + 1)^{9} + \frac{0}{10 !} (x + 1)^{10} + \dots

Refine

= \dots

f (x) = - \frac{12}{( x - 6 ) ^{2}}

\int \frac{5 x ^{3} + 3 x}{x - 1} d x

\int (x^{5} + 1) d x

y^{^{''}} + 8 y^{^{'}} - 9 y = e^{3 x}

\frac{d}{d x} (x (x - 21)^{2})