What is maclaurin t^2e^{2t} ?

The answer to maclaurin t^2e^{2t} is t^2+2t^3+2t^4+4/3 t^5+2/3 t^6+\ldots

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maclaurin t^2e^{2t}

maclaurin

t^{2} e^{2 t}

t^{2} + 2 t^{3} + 2 t^{4} + \frac{4}{3} t^{5} + \frac{2}{3} t^{6} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d t} ( t ^{2} e ^{2 t} ) ( 0 )}{1 !} t + \frac{\frac{d ^{2}}{d t ^{2}} ( t ^{2} e ^{2 t} ) ( 0 )}{2 !} t^{2} + \frac{\frac{d ^{3}}{d t ^{3}} ( t ^{2} e ^{2 t} ) ( 0 )}{3 !} t^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} t + \frac{2}{2 !} t^{2} + \frac{12}{3 !} t^{3} + \frac{48}{4 !} t^{4} + \frac{160}{5 !} t^{5} + \frac{480}{6 !} t^{6} + \dots

Refine

= t^{2} + 2 t^{3} + 2 t^{4} + \frac{4}{3} t^{5} + \frac{2}{3} t^{6} + \dots

\int_{- 7}^{7} x 49 - x^{2} d x

d y - d x + y = e^{4 x}

\int_{- 2}^{3} \frac{49}{x ^{4}} d x

\int 2 tan^{3} (x) sec (x) d x

f (x) = \frac{6 x}{x + 1}

What is maclaurin t^2e^{2t} ?
The answer to maclaurin t^2e^{2t} is t^2+2t^3+2t^4+4/3 t^5+2/3 t^6+\ldots