Question

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

Answer 1

Hint: 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
\[\begin{align}
  & \Rightarrow \left( 1+x \right){{\left\{ \left( 1+x \right)\left( 1-x+{{x}^{2}} \right) \right\}}^{40}} \\
 & \Rightarrow \left( 1+x \right){{\left\{ \left( 1-x+{{x}^{2}} \right)+x\left( 1-x+{{x}^{2}} \right) \right\}}^{40}} \\
 & \Rightarrow \left( 1+x \right){{\left( 1-x+{{x}^{2}}+x-{{x}^{2}}+{{x}^{3}} \right)}^{40}} \\
 & \Rightarrow \left( 1+x \right){{\left( 1+{{x}^{3}} \right)}^{40}} \\
\end{align}\]
\[\Rightarrow {{\left( 1+{{x}^{3}} \right)}^{40}}+x{{\left( 1+{{x}^{3}} \right)}^{40}}\] ………………………………………………(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.