Question

Expand the following expression in ascending powers of ‘x’ as far as ${{x}^{3}}$.
$\dfrac{1}{1+ax-a{{x}^{2}}-{{x}^{3}}}$

Answer 1

Hint: Suppose the expansion of the given term as $\left( {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+..... \right)$and compare the terms of variable ‘x’ in both sides of the equation to get number of equations. Solve them to get the values of${{a}_{0}},{{a}_{1}},{{a}_{2}},{{a}_{3}}$. Don’t write the higher power terms as it is not required.

Complete step-by-step answer:

Given expression in the problem is
$\dfrac{1}{1+ax-a{{x}^{2}}-{{x}^{3}}}........(i)$
As, we need to expand the given expression in equation (i) to the powers of ‘x’. So, let us assume the expansion of the expression as
$\dfrac{1}{1+ax-a{{x}^{2}}-{{x}^{3}}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+.....$
On cross-multiplying the above equation, we get
$1=\left( 1+ax-a{{x}^{2}}-{{x}^{3}} \right)\left( {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}....... \right)$
Now, let us multiply the brackets in Right-hand side, only upto the , as coefficient of higher powers is not asked, so we get
$1=\left( {{a}_{0}}+{{a}_{0}}ax-{{a}_{0}}a{{x}^{2}}-{{a}_{0}}{{x}^{3}} \right)+\left( {{a}_{1}}x+a{{a}_{1}}{{x}^{2}}-a{{a}_{1}}{{x}^{3}} \right)+\left( {{a}_{2}}{{x}^{2}}+a{{a}_{2}}{{x}^{3}} \right)+\left( {{a}_{3}}{{x}^{3}} \right)+....$
Now, taking the same powers of x in bracket, we get

& 1={{a}_{0}}+{{a}_{0}}ax+{{a}_{1}}x-{{a}_{0}}a{{x}^{2}}+a{{a}_{1}}{{x}^{2}}-{{a}_{2}}{{x}^{2}}-{{a}_{0}}{{x}^{3}}-a{{a}_{1}}{{x}^{3}}+a{{a}_{2}}{{x}^{3}}+{{a}_{3}}{{x}^{3}}+...... \\\ & \Rightarrow 1={{a}_{0}}+x\left( {{a}_{0}}a+{{a}_{1}} \right)+{{x}^{2}}\left( -{{a}_{0}}a+a{{a}_{1}}+{{a}_{2}} \right)+{{x}^{3}}\left( -{{a}_{0}}-a{{a}_{1}}+a{{a}_{2}}+{{a}_{3}} \right)+....... \\\ \end{aligned}$$ Now, we can write the left hand side of the above equation a $$1.{{x}^{0}}+0.x+0.{{x}^{2}}+0.{{x}^{3}}={{a}_{0}}+x\left( {{a}_{0}}a+{{a}_{1}} \right)+{{x}^{2}}\left( -{{a}_{0}}a+a{{a}_{1}}+{{a}_{2}} \right)+{{x}^{3}}\left( -{{a}_{0}}-a{{a}_{1}}+a{{a}_{2}}+{{a}_{3}} \right)...........$$ Now, compare the left hand side and right hand side of the above equation and get equations as $$\begin{aligned} & {{a}_{0}}=1.......(ii) \\\ & {{a}_{0}}a+{{a}_{1}}=0......(iii) \\\ & -{{a}_{0}}a+a{{a}_{1}}+{{a}_{2}}=0......(iv) \\\ & -{{a}_{0}}-a{{a}_{1}}+a{{a}_{2}}+{{a}_{3}}=0.......(v) \\\ \end{aligned}$$ So, we get value of as $${{a}_{0}}=1$$ Put value of $${{a}_{0}}$$from equation (ii) to equation (iii) as $$\begin{aligned} & 1(a)+{{a}_{1}}=0 \\\ & {{a}_{1}}=-a.......(vi) \\\ \end{aligned}$$ Now, put value of$${{a}_{0}}{{a}_{1}}$$to the equation (iv) to get value of $${{a}_{2}}$$as $$\begin{aligned} & -1(a)+a(-a)+{{a}_{2}}=0 \\\ & -a-{{a}^{_{2}}}+{{a}_{2}}=0 \\\ & {{a}_{2}}={{a}^{_{2}}}+a.......(vii) \\\ \end{aligned}$$ Now, we can put value of $${{a}_{0}},{{a}_{1}},{{a}_{2}}$$to equation (v) as $$\begin{aligned} & -1-a(-a)+a({{a}^{2}}+a)+{{a}_{3}}=0 \\\ & -1+{{a}^{2}}+{{a}^{3}}+{{a}^{2}}+{{a}_{3}}=0 \\\ & {{a}_{3}}=1-2{{a}^{2}}-{{a}^{3}}........(viii) \\\ \end{aligned}$$ Hence, we can write the expansion of the given expression as $$\dfrac{1}{1+a{{x}^{2}}-a{{x}^{2}}-{{x}^{3}}}=1-ax+({{a}^{2}}+a){{x}^{2}}+(1-2{{a}^{2}}-{{a}^{3}}){{x}^{3}}+..........$$ Note: One may use the binomial expansion of any index given as $${{(1+x)}^{n}}=1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)(n-2)}{3!}{{x}^{3}}+.........$$ It can be applied as $${{\left( 1+ax-a{{x}^{2}}-{{x}^{3}} \right)}^{-1}}=1+\left( -1 \right)\left( ax-a{{x}^{2}}-{{x}^{3}} \right)+\dfrac{\left( -1 \right)\left( -1-1 \right)}{2!}{{\left( ax-a{{x}^{2}}-{{x}^{3}} \right)}^{2}}+\dfrac{\left( -1 \right)\left( -1-1 \right)\left( -1-2 \right)}{3!}{{\left( ax-a{{x}^{2}}-{{x}^{3}} \right)}^{3}}+......$$ Now, use the above approach as well to expand the given series. Calculation is the key point of the question. We don’t need to take the terms with higher powers than$${{x}^{3}}$$.