Question

The coefficient of ${{x}^{50}}$ in ${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$ is

Answer 1

First of all, write ${{\left( 1+x \right)}^{41}}$ as the product of $\left( 1+x \right)$ and ${{\left( 1+x \right)}^{40}}$ . Our expression will look like $\left( 1+x \right){{\left( 1+x \right)}^{40}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$ . Now, modify the given expression as ${{\left( 1+{{x}^{3}} \right)}^{40}}+x{{\left( 1+{{x}^{3}} \right)}^{40}}$ . The coefficient of ${{x}^{50}}$ is equal to the summation of the coefficients of ${{x}^{50}}$ from ${{\left( 1+{{x}^{3}} \right)}^{40}}$ and $x{{\left( 1+{{x}^{3}} \right)}^{40}}$ . Expand the expression ${{\left( 1+{{x}^{3}} \right)}^{40}}$ by using the binomial expansion formula, ${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+..........{{+}^{n}}{{C}_{n}}{{x}^{n}}$ . Now, we can see that the exponent of x is a multiple of 3. But we have to find the coefficient of ${{x}^{50}}$ and 50 is not divisible by 3. So, the coefficient of ${{x}^{50}}$ in ${{\left( 1+{{x}^{3}} \right)}^{40}}$ is zero. For the coefficient of ${{x}^{50}}$ in $x{{\left( 1+{{x}^{3}} \right)}^{40}}$ we have to find the coefficient of ${{x}^{49}}$ in the expression ${{\left( 1+{{x}^{3}} \right)}^{40}}$ . But in the expansion of ${{\left( 1+{{x}^{3}} \right)}^{40}}$ , we can see that the exponent of x is a multiple of 3 and 49 is not divisible by 3. So, the coefficient of ${{x}^{49}}$ is also zero. Now, solve it further and get the value of the coefficient of ${{x}^{50}}$ in ${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$.

Complete step-by-step solution:
According to the question, we are given an expression and we have to find the coefficient of ${{x}^{50}}$.
The given expression = ${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$ ………………………………………………(1)
We can write ${{\left( 1+x \right)}^{41}}$ as the product of $\left( 1+x \right)$ and ${{\left( 1+x \right)}^{40}}$ , that is
${{\left( 1+x \right)}^{41}}=\left( 1+x \right)\times {{\left( 1+x \right)}^{40}}$ ………………………………………….(2)
Now, from equation (1) and equation (2), we get
$\Rightarrow {{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$
$\Rightarrow \left( 1+x \right){{\left( 1+x \right)}^{40}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$ ………………………………………….(3)
On simplifying equation (3), we get

& \Rightarrow \left( 1+x \right){{\left\\{ \left( 1+x \right)\left( 1-x+{{x}^{2}} \right) \right\\}}^{40}} \\\ & \Rightarrow \left( 1+x \right){{\left\\{ \left( 1-x+{{x}^{2}} \right)+x\left( 1-x+{{x}^{2}} \right) \right\\}}^{40}} \\\ & \Rightarrow \left( 1+x \right){{\left( 1-x+{{x}^{2}}+x-{{x}^{2}}+{{x}^{3}} \right)}^{40}} \\\ & \Rightarrow \left( 1+x \right){{\left( 1+{{x}^{3}} \right)}^{40}} \\\ \end{aligned}$$ $$\Rightarrow {{\left( 1+{{x}^{3}} \right)}^{40}}+x{{\left( 1+{{x}^{3}} \right)}^{40}}$$ ………………………………………………(4) We are asked to find the coefficient of $${{x}^{50}}$$ in the above expression. For the coefficient of $${{x}^{50}}$$ , we have to find the summation of the coefficients of $${{x}^{50}}$$ from $${{\left( 1+{{x}^{3}} \right)}^{40}}$$ and $$x{{\left( 1+{{x}^{3}} \right)}^{40}}$$ …………………………..(5) Let us find the coefficient of $${{x}^{50}}$$ from the expression $${{\left( 1+{{x}^{3}} \right)}^{40}}$$ . We know the formula for binomial expansion, $${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+..........{{+}^{n}}{{C}_{n}}{{x}^{n}}$$ ………………………………………(6) Using the formula shown in equation (6) and applying it for the expansion of $${{\left( 1+{{x}^{3}} \right)}^{40}}$$ , we get $${{\left( 1+{{x}^{3}} \right)}^{40}}{{=}^{40}}{{C}_{0}}{{+}^{40}}{{C}_{1}}\left( {{x}^{3}} \right){{+}^{40}}{{C}_{2}}{{\left( {{x}^{3}} \right)}^{2}}+..........{{+}^{40}}{{C}_{40}}{{\left( {{x}^{3}} \right)}^{40}}$$ ………………………………………(7) In equation (7), we can see that the exponent of x is a multiple of 3. But we have to find the coefficient of $${{x}^{50}}$$ and 50 is not divisible by 3. So, the coefficient of $${{x}^{50}}$$ in $${{\left( 1+{{x}^{3}} \right)}^{40}}$$ is zero ………………………………………..(8) To find the coefficient of $${{x}^{50}}$$ in $$x{{\left( 1+{{x}^{3}} \right)}^{40}}$$ we have to find the coefficient of $${{x}^{49}}$$ in the expression $${{\left( 1+{{x}^{3}} \right)}^{40}}$$ . From equation (7), we have the expression for the expansion of $${{\left( 1+{{x}^{3}} \right)}^{40}}$$ . Similarly, in equation (7), we can see that the exponent of x is a multiple of 3. But we have to find the coefficient of $${{x}^{49}}$$ and 49 is not divisible by 3. So, the coefficient of $${{x}^{50}}$$ in $$x{{\left( 1+{{x}^{3}} \right)}^{40}}$$ is also zero ………………………………………..(9) Now, from equation (5), equation (8), and equation (9), we have Coefficient of $${{x}^{50}}$$ in $${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$$ = Coefficient of $${{x}^{50}}$$ in $${{\left( 1+{{x}^{3}} \right)}^{40}}$$ + Coefficient of $${{x}^{50}}$$ in $$x{{\left( 1+{{x}^{3}} \right)}^{40}}$$ $$\Rightarrow $$ Coefficient of $${{x}^{50}}$$ in $${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$$ = 0 + 0 $$\Rightarrow $$ Coefficient of $${{x}^{50}}$$ in $${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$$ = 0 …………………………………………(10) Therefore, the coefficient of $${{x}^{50}}$$ in $${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$$ is equal to zero. **Note:** In this question, one might try to simply expand the expression $${{\left( 1+x \right)}^{41}}{{\left( 1-x+{{x}^{2}} \right)}^{40}}$$ by using the binomial expansion formula, $${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+..........{{+}^{n}}{{C}_{n}}{{x}^{n}}$$ . Doing so would lead to complexity and so with the calculation mistakes.