Question

Number of terms in the expansion of ${{\left( 1-x \right)}^{51}}{{\left( 1+x+{{x}^{2}} \right)}^{50}}$ is
a)50
b)51
c)100
d)102

Answer 1

Hint: Here, the expansion ${{\left( 1-x \right)}^{51}}{{\left( 1+x+{{x}^{2}} \right)}^{50}}$ is given and we have to find total number of terms after expanding. So, we will use the Binomial theorem that states the ${{n}^{th}}$ power of $\left( a+b \right)$ can be used as sum of $n+1$ terms, where $n>0$ .

Complete step-by-step answer:
Here, the expansion is given i.e. ${{\left( 1-x \right)}^{51}}{{\left( 1+x+{{x}^{2}} \right)}^{50}}$ and we will use Binomial theorem.
Using Binomial theorem, which is
$\begin{align}
  & {{\left( a+b \right)}^{n}}={{a}^{n}}+\left( {}^{n}{{C}_{1}} \right){{a}^{n-1}}b+\left( {}^{n}{{C}_{1}} \right){{a}^{n-1}}b+\left( {}^{n}{{C}_{2}} \right){{a}^{n-2}}{{b}^{2}}+.......+\left( {}^{n}{{C}_{n-1}} \right)a{{b}^{n-1}}+{{b}^{n}} \\
 & \\
\end{align}$ ………………(i)
On solving the given expression, we get
${{\left( 1-x \right)}^{51}}\cdot {{\left( +x+{{x}^{2}} \right)}^{50}}=\left( 1-x \right){{\left[ \left( 1-x \right)\left( 1+x+{{x}^{2}} \right) \right]}^{50}}$ …………………………………(ii)
Taking $\left( 1-x \right)$ term common to make power equal.
Now, on multiplying $\left( 1-x \right)\left( 1+x+{{x}^{2}} \right)$ we get,
$=1+x+{{x}^{2}}-x-{{x}^{2}}-{{x}^{3}}$
$=1-{{x}^{3}}$ ………………………….(iii)
Putting equation (iii) in (ii), we get
${{\left( 1-x \right)}^{51}}\cdot {{\left( +x+{{x}^{2}} \right)}^{50}}=\left( 1-x \right){{\left( 1-{{x}^{3}} \right)}^{50}}$ ………………………………………(iv)
Now using theorem on ${{\left( 1-{{x}^{3}} \right)}^{50}}$ where $a=1,b=\left( -{{x}^{3}} \right),n=50$
${{\left( 1-{{x}^{3}} \right)}^{50}}=1+\left( {}^{50}{{C}_{1}} \right){{\left( 1 \right)}^{49}}\left( -{{x}^{3}} \right)+\left( {}^{50}{{C}_{2}} \right){{\left( 1 \right)}^{48}}{{\left( -{{x}^{3}} \right)}^{2}}+.....+\left( {}^{50}{{C}_{49}} \right)\left( 1 \right){{\left( -{{x}^{3}} \right)}^{49}}+\left( {}^{50}{{C}_{50}} \right){{\left( -{{x}^{3}} \right)}^{50}}$
${{\left( 1-{{x}^{3}} \right)}^{50}}=1+\left( {}^{50}{{C}_{1}} \right)\left( -{{x}^{3}} \right)+\left( {}^{50}{{C}_{2}} \right){{\left( 1 \right)}^{48}}\left( {{x}^{6}} \right)+.....+\left( {}^{50}{{C}_{49}} \right)\left( 1 \right)\left( -{{x}^{147}} \right)+\left( {}^{50}{{C}_{50}} \right)\left( {{x}^{150}} \right)$ ….….(v)
Here considering ${}^{n}{{C}_{r}}$ term as ${{a}_{r}}$ and substituting this in equation (v), we get
${{\left( 1-{{x}^{3}} \right)}^{50}}=1+{{a}_{1}}\left( -{{x}^{3}} \right)+{{a}_{2}}{{\left( 1 \right)}^{48}}\left( {{x}^{6}} \right)+.....+{{a}_{49}}\left( 1 \right)\left( -{{x}^{147}} \right)+{{a}_{50}}\left( {{x}^{150}} \right)$
Now, putting the above expansion in equation (iv) and on solving we get,
$=\left( 1-x \right)\left[ 1-{{a}_{1}}{{x}^{3}}+{{a}_{2}}{{x}^{6}}+.....-{{a}_{49}}{{x}^{147}}+{{a}_{50}}{{x}^{150}} \right]$
Multiplying both the brackets, we get
$=\left( 1-{{a}_{1}}{{x}^{3}}+{{a}_{2}}{{x}^{6}}-{{a}_{3}}{{x}^{9}}.....{{a}_{50}}{{x}^{150}} \right)-\left( x-{{a}_{1}}{{x}^{4}}+{{a}_{2}}{{x}^{7}}-{{a}_{3}}{{x}^{10}}+....{{a}_{50}}{{x}^{151}} \right)$
$=1-x-{{a}_{1}}{{x}^{3}}+{{a}_{1}}{{x}^{4}}+{{a}_{2}}{{x}^{6}}+{{a}_{2}}{{x}^{7}}+....-{{a}_{50}}{{x}^{151}}$
So, in both the brackets there are 50 terms plus additional terms 1 and $\left( -x \right)$ in each bracket respectively.
So, total 51 terms are there in each bracket after expanding.
Total number of terms $=2\left( 51 \right)=102$
Hence, option (d) is correct.

Note: Students should be very careful while expanding the expression as there are chances of sign changing while multiplying the brackets. Also, the formula of the binomial theorem should be known properly.