# Expand the quantity sqrt[5]{P + t}| about 0 in terms of frac{t}{P}| Give four non-zero terms.

## Question:

Expand the quantity

{eq}\sqrt[5]{P + t}| {/eq}

about {eq}0 {/eq} in terms of {eq}\left.\begin{matrix} \frac{t}{P} \end{matrix}\right| {/eq}

Give four non-zero terms.

## Binomial theorem

The binomial theorem is used to find the complex mathematical expression of nth power. The binomial theorem is used to evaluate the extension of a series. It is also used in the probability.

Given

• First term of the quantity is {eq}a = P {/eq}
• Second term of the quantity is {eq}b = t {/eq}

The expression for the binomial expansion,

{eq}{\left( {a + b} \right)^n} = 1 + nx + n\dfrac{{\left( {n - 1} \right)}}{{2!}}{x^2} + n\dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3}.....\;......\left( 1 \right) {/eq}

The expression for the given quantity,

{eq}{\left( {P + t} \right)^{\dfrac{1}{5}}} = {P^{\dfrac{1}{5}}}{\left( {1 + \dfrac{t}{P}} \right)^{\dfrac{1}{5}}} {/eq}

Substitute the value of {eq}a = P,\;b = t\;and\;n = \dfrac{1}{5} {/eq} in equation (1),

{eq}\begin{align*} {\left( {P + t} \right)^{\dfrac{1}{5}}} &= {P^{\dfrac{1}{5}}}\left[ {1 + \dfrac{1}{5}\left( {\dfrac{t}{P}} \right) + \dfrac{{\dfrac{1}{5}\left( {\dfrac{1}{5} - 1} \right)}}{{2!}}{{\left( {\dfrac{t}{P}} \right)}^2} + \dfrac{{\dfrac{1}{5}\left( {\dfrac{1}{5} - 1} \right)\left( {\dfrac{1}{5} - 2} \right)}}{{3!}}{x^3} + ....} \right]\\ {\left( {P + t} \right)^{\dfrac{1}{5}}} &= {P^{\dfrac{1}{5}}}\left[ {1 + \dfrac{1}{5} \cdot \dfrac{t}{P} - \dfrac{2}{{25}} \cdot {{\left( {\dfrac{t}{p}} \right)}^2} + \dfrac{6}{{125}}{{\left( {\dfrac{t}{P}} \right)}^3} - ....} \right] \end{align*} {/eq}

Therefore, expansion of the expression is {eq}{P^{\dfrac{1}{5}}}\left[ {1 + \dfrac{1}{5} \cdot \dfrac{t}{P} - \dfrac{2}{{25}} \cdot {{\left( {\dfrac{t}{p}} \right)}^2} + \dfrac{6}{{125}}{{\left( {\dfrac{t}{P}} \right)}^3} - ....} \right] {/eq}