Question

The coefficient of ${{x}^{65}}$ in the expansion of A.${{(1+x)}^{131}}{{({{x}^{2}}-x+1)}^{130}}$ is
B.${}^{130}{{C}_{65}}+{}^{129}{{C}_{66}}$
C.${}^{130}{{C}_{65}}+{}^{129}{{C}_{55}}$
D.${}^{130}{{C}_{66}}+{}^{129}{{C}_{65}}$
E.None of these

Answer 1

Here we have to the coefficient of ${{x}^{65}}$ in the expansion of ${{(1+x)}^{131}}{{({{x}^{2}}-x+1)}^{130}}$ . First simplify the expansion and use ${{\left( 1\text{ }+\text{ }x \right)}^{n}}~=\text{ }{{a}_{0~}}+\text{ }{{a}_{1}}~x\text{ }+\text{ }{{a}_{2}}~{{x}^{2~}}+\text{ }\ldots \text{ }+{{a}_{r}}~{{x}^{r~}}+...$.

Complete step-by-step answer:
Now let us take the expansion ${{(1+x)}^{131}}{{({{x}^{2}}-x+1)}^{130}}$.
Simplifying the above expansion we get,
${{(1+x)}^{131}}{{({{x}^{2}}-x+1)}^{130}}=\left( 1+x \right){{(1+x)}^{130}}{{({{x}^{2}}-x+1)}^{130}}$
We know the property that, ${{a}^{b}}{{c}^{b}}={{(a\times c)}^{b}}$.
So, using the above property we get,
${{(1+x)}^{131}}{{({{x}^{2}}-x+1)}^{130}}=\left( 1+x \right){{\left( (1+x)({{x}^{2}}-x+1) \right)}^{130}}$ …………… (1)
Now let us take ${{\left( (1+x)({{x}^{2}}-x+1) \right)}^{130}}$ and simplify it.
${{\left( (1+x)({{x}^{2}}-x+1) \right)}^{130}}={{\left( {{x}^{2}}-x+1+{{x}^{3}}-{{x}^{2}}+x \right)}^{130}}$
Again, simplifying we get,
${{\left( (1+x)({{x}^{2}}-x+1) \right)}^{130}}={{\left( 1+{{x}^{3}} \right)}^{130}}$ ……….. (2)
Now let us substitute (2) in (1) we get,
${{(1+x)}^{131}}{{({{x}^{2}}-x+1)}^{130}}=\left( 1+x \right){{\left( 1+{{x}^{3}} \right)}^{130}}$ ……………. (3)
Now we know the formula that, ${{\left( 1\text{ }+\text{ }x \right)}^{n}}~=\text{ }{{a}_{0~}}+\text{ }{{a}_{1}}~x\text{ }+\text{ }{{a}_{2}}~{{x}^{2~}}+\text{ }\ldots \text{ }+{{a}_{r}}~{{x}^{r~}}+...$.
So, using the above formula for ${{\left( 1+{{x}^{3}} \right)}^{130}}$.
${{\left( 1+{{x}^{3}} \right)}^{130}}=(1+{{a}_{1}}{{x}^{3}}+{{a}_{2}}{{\left( {{x}^{3}} \right)}^{2}}+{{a}_{3}}{{\left( {{x}^{3}} \right)}^{3}}+....+{{a}_{21}}{{\left( {{x}^{3}} \right)}^{21}}+{{a}_{22}}{{\left( {{x}^{3}} \right)}^{22}}+....+{{a}_{130}}{{\left( {{x}^{3}} \right)}^{130}})$
Now simplifying we get,
${{\left( 1+{{x}^{3}} \right)}^{130}}=(1+{{a}_{1}}{{x}^{3}}+{{a}_{2}}\left( {{x}^{6}} \right)+{{a}_{3}}\left( {{x}^{9}} \right)+....+{{a}_{21}}\left( {{x}^{63}} \right)+{{a}_{22}}\left( {{x}^{66}} \right)+....+{{a}_{130}}\left( {{x}^{390}} \right))$
Here, substituting ${{\left( 1+{{x}^{3}} \right)}^{130}}$ in (3) we get,
$\left( 1+x \right){{\left( 1+{{x}^{3}} \right)}^{130}}=\left( 1+x \right)(1+{{a}_{1}}{{x}^{3}}+{{a}_{2}}\left( {{x}^{6}} \right)+{{a}_{3}}\left( {{x}^{9}} \right)+....+{{a}_{21}}\left( {{x}^{63}} \right)+{{a}_{22}}\left( {{x}^{66}} \right)+....+{{a}_{130}}\left( {{x}^{390}} \right))$
In above we can see that, coefficient of ${{x}^{65}}$ term is not present.
So, the coefficient of ${{x}^{65}}$ is $0$.
Therefore, the correct answer is option (E).

Additional information:
Binomial Theorem - As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of Binomial. The Binomial Theorem is the method of expanding an expression which has been raised to any finite power. A binomial Theorem is a powerful tool of expansion, which has application in Algebra, probability, etc. Binomial coefficients refer to the integers which are coefficients in the binomial theorem. The Binomial Expansion Theorem is an algebra formula that describes the algebraic expansion of powers of a binomial. According to the binomial expansion theorem, it is possible to expand any power of $x+y$ into a sum of the terms.

Note: We can see that the term ${{x}^{65}}$ is not present, so its coefficient is $0$. We have used the formula ${{\left( 1\text{ }+\text{ }x \right)}^{n}}~=\text{ }{{a}_{0~}}+\text{ }{{a}_{1}}~x\text{ }+\text{ }{{a}_{2}}~{{x}^{2~}}+\text{ }\ldots \text{ }+{{a}_{r}}~{{x}^{r~}}+...$. Also, used some properties due to which the problem could be simplified.