Question

If the sum of coefficients in expansion of ${{({{\alpha }^{2}}{{x}^{2}}+2\alpha x+1)}^{51}}$ is zero, then $\alpha$ is
A. 1
B. -1
C. 0
D. None of these

Answer 1

Hint: Use $x=1$ and ${{(x+a)}^{n}}$ . Then simplify it and keep it equal to $0$. This will give the value of $\alpha$ as required in the question.

Complete step by step answer:
In question it is given that the sum of coefficients of expansion ${{({{\alpha }^{2}}{{x}^{2}}+2\alpha x+1)}^{51}}$ is 0.
So in sum of coefficients of any expansion to solve we put $x=1$ ,
So we have an expansion ${{({{\alpha }^{2}}{{x}^{2}}+2\alpha x+1)}^{51}}$ ,
So putting $x=1$ in above expansion we get the expansion as,
$\begin{aligned} & {{({{\alpha }^{2}}{{x}^{2}}+2\alpha x+1)}^{51}}={{({{\alpha }^{2}}{{(1)}^{2}}+2\alpha (1)+1)}^{51}} \\\ & ={{({{\alpha }^{2}}+2\alpha +1)}^{51}} \\\ \end{aligned}$
So we get final simplification as ${{({{\alpha }^{2}}+2\alpha +1)}^{51}}$ ,
As we know the identity ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ ,
So we can write $({{\alpha }^{2}}+2\alpha +1)={{(\alpha +1)}^{2}}$ ……….. (from above identity)
So ${{({{\alpha }^{2}}+2\alpha +1)}^{51}}$ becomes,
${{({{\alpha }^{2}}+2\alpha +1)}^{51}}={{(\alpha +1)}^{102}}$ ,
The sum of the coefficients of the terms in the expansion of a binomial raised to a power cannot be determined beforehand, taking a simple example below,
${{(x+1)}^{2}}={{x}^{2}}+2x+1,\sum{{{C}_{x}}=4}$
This is because of the second term of the binomial which is a constant. This constant will also contribute to the coefficients of the terms.
According to the binomial theorem, the ${{(r+1)}^{th}}$ term in the expansion of ${{(x+a)}^{n}}$ is,
${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}$
You can see the ${}^{n}{{C}_{r}}$ being used here which is the binomial coefficient. The sum of the binomial coefficients will be ${{2}^{n}}$ because, as we know that,
$\sum\nolimits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right)}={{2}^{n}}$
The middle term depends upon the value of $n$ ,
If $n$ is even: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (odd).
If $n$ is odd: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (even).

If $n$ is positive integer,
${{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}$
So replacing $a$ by $1$ and $b$ by $x$we get,
${{(1+x)}^{n}}=={}^{n}{{C}_{0}}{{\left( x \right)}^{0}}+{}^{n}{{C}_{1}}{{\left( x \right)}^{1}}+{}^{n}{{C}_{2}}{{\left( x \right)}^{2}}+{}^{n}{{C}_{3}}{{\left( x \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{\left( x \right)}^{n}}$
So expanding ${{(\alpha +1)}^{102}}$ using binomial theorem, we get,
So applying above theorem we get,
${{(\alpha +1)}^{102}}={}^{102}{{C}_{0}}{{\left( \alpha \right)}^{0}}+{}^{102}{{C}_{1}}{{\left( \alpha \right)}^{1}}+{}^{102}{{C}_{2}}{{\left( \alpha \right)}^{2}}+{}^{102}{{C}_{3}}{{\left( \alpha \right)}^{3}}+...........+{}^{102}{{C}_{102}}{{\left( \alpha \right)}^{102}}$
So simplifying in simple manner,
${{(\alpha +1)}^{102}}={}^{102}{{C}_{0}}+{}^{102}{{C}_{1}}{{\left( \alpha \right)}^{1}}+{}^{102}{{C}_{2}}{{\left( \alpha \right)}^{2}}+{}^{102}{{C}_{3}}{{\left( \alpha \right)}^{3}}+...........+{}^{102}{{C}_{102}}{{\left( \alpha \right)}^{102}}$ …………. (1)
It is given in question that,
${}^{102}{{C}_{0}}+{}^{102}{{C}_{1}}{{\left( \alpha \right)}^{1}}+{}^{102}{{C}_{2}}{{\left( \alpha \right)}^{2}}+{}^{102}{{C}_{3}}{{\left( \alpha \right)}^{3}}+...........+{}^{102}{{C}_{102}}{{\left( \alpha \right)}^{102}}=0$ …………… (2)
So from (1) and (2),
${{(\alpha +1)}^{102}}={}^{102}{{C}_{0}}+{}^{102}{{C}_{1}}{{\left( \alpha \right)}^{1}}+{}^{102}{{C}_{2}}{{\left( \alpha \right)}^{2}}+{}^{102}{{C}_{3}}{{\left( \alpha \right)}^{3}}+...........+{}^{102}{{C}_{102}}{{\left( \alpha \right)}^{102}}=0$
So we get,
${{(\alpha +1)}^{102}}=0$
So simplifying in simple manner we get,
$(\alpha +1)=0$
$\alpha =-1$
Therefore we get the final value of $\alpha$ as $-1$.

The correct answer is option (B).

Note: Read the question properly. Be familiar with the formula ${{(x+a)}^{n}}$ . Don’t confuse while simplifying and don’t jumble between $n$ and $r$. Use the formula${{(1+x)}^{n}}=={}^{n}{{C}_{0}}{{\left( x \right)}^{0}}+{}^{n}{{C}_{1}}{{\left( x \right)}^{1}}+{}^{n}{{C}_{2}}{{\left( x \right)}^{2}}+{}^{n}{{C}_{3}}{{\left( x \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{\left( x \right)}^{n}}$. Expand ${{(\alpha +1)}^{102}}$ in a proper manner.