Question

Prove that the coefficient of ${{\text{x}}^4}$ in ${({\text{1 + x - 2}}{{\text{x}}^2})^6}$ is -45 and if the complete expansion of the expression is ${\text{1 + }}{{\text{a}}_1}{\text{x + }}{\text{ }}{{\text{a}}_2}{{\text{x}}^2}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}{{\text{x}}^{12}}$. Prove that ${{\text{a}}_2}{\text{ + }}{\text{ }}{{\text{a}}_4}{\text{ + }}{{\text{a}}_6}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}{\text{ = 31}}$.

Answer 1

Hint: To solve this question, we will use the multinomial theorem. The formula to find a term in the using multinomial theorem is ${{\text{T}}_{\text{n}}}{\text{ = }}\dfrac{{6!}}{{{\text{a!b!c!}}}}{(1)^{\text{a}}}{({\text{x)}}^{\text{b}}}{( - 2{{\text{x}}^2})^{\text{c}}}$.

Complete step-by-step solution:

Now, we are given ${({\text{1 + x - 2}}{{\text{x}}^2})^6}$. We have to find the coefficient of ${{\text{x}}^4}$ in the given expansion. To find the coefficient, we will use a multinomial theorem. Multinomial theorem is used to find terms in multinomials instead of binomials.
The formula to find a term in the given expansion is
${{\text{T}}_{\text{n}}}{\text{ = }}\dfrac{{6!}}{{{\text{a!b!c!}}}}{(1)^{\text{a}}}{({\text{x)}}^{\text{b}}}{( - 2{{\text{x}}^2})^{\text{c}}}$, where a, b, c are the powers of 1, x, $2{{\text{x}}^2}$ present in the term we are going to find.
In the above formula, a + b + c = 6.
Now, we have to find the coefficient of ${{\text{x}}^4}$. So, we will find the term having ${{\text{x}}^4}$. For this, from the formula we can clearly see that there are two terms involving x, i.e. x and $2{{\text{x}}^2}$ and their powers are b and c respectively.
So, to get ${{\text{x}}^4}$, we have b + 2c = 4. So, following combinations are possible:
When b = 0, we get c = 2. Therefore, a = 4
When b = 2, we get c = 1. Therefore, a = 3
When b = 4, we get c = 0. Therefore, a = 2
So, total 3 combinations are possible. Now, to find the coefficient of ${{\text{x}}^4}$, we will put the values of a, b, c in the formula of multinomial theorem and add all the terms we get.
Therefore, we can write as
Coefficient of ${{\text{x}}^4}$ = $\dfrac{{6!}}{{4!2!0!}}{(1)^4}{( - 2)^2}{\text{ + }}\dfrac{{6!}}{{3!2!1!}}{(1)^3}{( - 2)^1}{\text{ + }}\dfrac{{6!}}{{2!4!0!}}{(1)^2}{( - 2)^0}$
Therefore, on solving we get
Coefficient of ${{\text{x}}^4}$ = 60 - 120 + 15
So, the coefficient of ${{\text{x}}^4}$ = -45.
Now, we are given
${({\text{1 + x - 2}}{{\text{x}}^2})^6}$ = ${\text{1 + }}{{\text{a}}_1}{\text{x + }}{\text{ }}{{\text{a}}_2}{{\text{x}}^2}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}{{\text{x}}^{12}}$ …. (1)
Putting x = -1 in the above expression, we get
${({\text{1 - 1 - 2( - 1}}{{\text{)}}^2}{\text{)}}^6}$ = ${\text{1 + }}{{\text{a}}_1}{\text{( - 1) + }}{\text{ }}{{\text{a}}_2}{( - 1)^2}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}{( - 1)^{12}}$
64 = ${\text{1 - }}{{\text{a}}_1}{\text{ + }}{\text{ }}{{\text{a}}_2}{\text{ - }}{{\text{a}}_3}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}$ … (2)
Now, putting x = 1 in the equation (1), we get
${({\text{1 + 1 - 2(1}}{{\text{)}}^2}{\text{)}}^6}$ = ${\text{1 + }}{{\text{a}}_1}{\text{(1) + }}{\text{ }}{{\text{a}}_2}{(1)^2}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}{(1)^{12}}$
0 = ${\text{1 + }}{{\text{a}}_1}{\text{ + }}{\text{ }}{{\text{a}}_2}{\text{ + }}{{\text{a}}_3}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}$ … (3)
Adding, equation (2) and (3), we get
64 = 2 (${\text{1 + }}{{\text{a}}_2}{\text{ + }}{\text{ }}{{\text{a}}_4}{\text{ + }}{{\text{a}}_6}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}$)
32 = ${\text{1 + }}{{\text{a}}_2}{\text{ + }}{\text{ }}{{\text{a}}_4}{\text{ + }}{{\text{a}}_6}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}$
Therefore,
${{\text{a}}_2}{\text{ + }}{\text{ }}{{\text{a}}_4}{\text{ + }}{{\text{a}}_6}{\text{ + }}.........{\text{ + }}{{\text{a}}_{12}}{\text{ = 31}}$
Hence proved.

Note: When we come up with such types of questions, we have to apply a multinomial theorem to find the terms asked in the questions. This formula is applicable to every multinomial including binomial but we avoid using this formula in case of binomials as binomial has its own specific formula. While applying this formula, one thing to keep in mind is that when you find a term or coefficient asked in the variable, don’t make such conditions which makes you solve the factorial of a fraction. Because factorial of a fraction is not possible.