Evaluate $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=10$ and $n=11$?
Answer
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Hint: We start solving the problem by assigning the variable for $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$. We then use the fact that ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{o}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$ for ${{\left( 1+x \right)}^{10}}$, ${{\left( 1-x \right)}^{10}}$ for $n=10$ and multiply both of them. We do this step similarly for $n=11$. We then compare the coefficient of ${{x}^{10}}$ on both sides for $n=10$ and the coefficient of ${{x}^{11}}$ for $n=11$. We then use the fact that that the general form of terms in the binomial expansion ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$, where $0\le r\le n$ and make necessary calculations to get the required values.
Complete step-by-step solution:
According to the problem, we need to find the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=10$ and $n=11$.
Let us first find the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=10$. Let us assume this value is ‘y’.
So, we get $y=~{}^{10}C_{0}^{2}-{}^{10}C_{1}^{2}+{}^{10}C_{2}^{2}-{}^{10}C_{3}^{2}+{}^{10}C_{4}^{2}-{}^{10}C_{5}^{2}+{}^{10}C_{6}^{2}-{}^{10}C_{7}^{2}+{}^{10}C_{8}^{2}-{}^{10}C_{9}^{2}+{}^{10}C_{10}^{2}$ ---(1).
We know that ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{o}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$.
So, we get ${{\left( 1-x \right)}^{10}}={}^{10}{{C}_{o}}+{}^{10}{{C}_{1}}\left( -x \right)+{}^{10}{{C}_{2}}{{\left( -x \right)}^{2}}+{}^{10}{{C}_{3}}{{\left( -x \right)}^{3}}+......+{}^{10}{{C}_{10}}{{\left( -x \right)}^{10}}$.
$\Rightarrow {{\left( 1-x \right)}^{10}}={}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-{}^{10}{{C}_{3}}{{x}^{3}}+......+{}^{10}{{C}_{10}}{{x}^{10}}$ ---(2).
Now, we know that ${{\left( 1+x \right)}^{10}}={{\left( x+1 \right)}^{10}}={}^{10}{{C}_{o}}{{\left( x \right)}^{10}}+{}^{10}{{C}_{1}}{{\left( x \right)}^{9}}+{}^{10}{{C}_{2}}{{\left( x \right)}^{8}}+{}^{10}{{C}_{3}}{{\left( x \right)}^{7}}+......+{}^{10}{{C}_{10}}$.
$\Rightarrow {{\left( 1+x \right)}^{10}}={}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+{}^{10}{{C}_{3}}{{x}^{7}}+......+{}^{10}{{C}_{10}}$ ---(3).
Let us multiply equations (2) and (3).
We get \[\Rightarrow {{\left( 1-x \right)}^{10}}\times {{\left( 1+x \right)}^{10}}=\left( {}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-......+{}^{10}{{C}_{10}}{{x}^{10}} \right)\times \left( {}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+......+{}^{10}{{C}_{10}} \right)\].
\[\Rightarrow {{\left( \left( 1-x \right)\times \left( 1+x \right) \right)}^{10}}=\left( {}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-......+{}^{10}{{C}_{10}}{{x}^{10}} \right)\times \left( {}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+......+{}^{10}{{C}_{10}} \right)\].
\[\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{10}}=\left( {}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-......+{}^{10}{{C}_{10}}{{x}^{10}} \right)\times \left( {}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+......+{}^{10}{{C}_{10}} \right)\].
Now, let us compare the coefficients of ${{x}^{10}}$ on both sides.
Let us find the general form of terms in ${{\left( 1-{{x}^{2}} \right)}^{10}}$. We know that the general form of terms in the binomial expansion ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$, where $0\le r\le n$.
So, the general forms of terms in ${{\left( 1-{{x}^{2}} \right)}^{10}}$ is ${}^{10}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( {{x}^{2}} \right)}^{10-r}}={}^{10}{{C}_{r}}{{\left( x \right)}^{20-2r}}$, $0\le r\le 10$.
We need to find the coefficient of ${{x}^{10}}$. So, we get $20-2r=10$.
$\Rightarrow 2r=10$.
$\Rightarrow r=5$.
So, the coefficient of ${{x}^{10}}$ in ${{\left( 1-{{x}^{2}} \right)}^{10}}$ is ${}^{10}{{C}_{5}}$.
So, we get ${}^{10}{{C}_{5}}={}^{10}{{C}_{0}}\times {}^{10}{{C}_{0}}+{}^{10}{{C}_{1}}\times \left( -{}^{10}{{C}_{1}} \right)+{}^{10}{{C}_{2}}\times {}^{10}{{C}_{2}}+{}^{10}{{C}_{3}}\times \left( -{}^{10}{{C}_{3}} \right)+......+{}^{10}{{C}_{10}}\times {}^{10}{{C}_{10}}$.
$\Rightarrow {}^{10}{{C}_{5}}={}^{10}C_{0}^{2}-{}^{10}C_{1}^{2}+{}^{10}C_{2}^{3}-{}^{10}C_{3}^{2}+......+{}^{10}C_{10}^{2}$.
From equation (1), we get $y={}^{10}{{C}_{5}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
$\Rightarrow y=\dfrac{10!}{5!5!}$.
$\Rightarrow y=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}$.
$\Rightarrow y=2\times 9\times 2\times 7$.
$\Rightarrow y=252$.
So, the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=10$ is 252.
Now, let us find the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=11$. Let us assume this value is ‘z’.
So, we get $z=~{}^{11}C_{0}^{2}-{}^{11}C_{1}^{2}+{}^{11}C_{2}^{2}-{}^{11}C_{3}^{2}+{}^{11}C_{4}^{2}-{}^{11}C_{5}^{2}+{}^{11}C_{6}^{2}-{}^{11}C_{7}^{2}+{}^{11}C_{8}^{2}-{}^{11}C_{9}^{2}+{}^{11}C_{10}^{2}-{}^{11}C_{11}^{2}$ ---(4).
Now, we have ${{\left( 1-x \right)}^{11}}={}^{11}{{C}_{o}}+{}^{11}{{C}_{1}}\left( -x \right)+{}^{11}{{C}_{2}}{{\left( -x \right)}^{2}}+{}^{11}{{C}_{3}}{{\left( -x \right)}^{3}}+......+{}^{11}{{C}_{11}}{{\left( -x \right)}^{11}}$.
$\Rightarrow {{\left( 1-x \right)}^{11}}={}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-{}^{11}{{C}_{3}}{{x}^{3}}+......-{}^{11}{{C}_{11}}{{x}^{11}}$ ---(5).
Now, we know that ${{\left( 1+x \right)}^{11}}={{\left( x+1 \right)}^{11}}={}^{11}{{C}_{o}}{{\left( x \right)}^{11}}+{}^{11}{{C}_{1}}{{\left( x \right)}^{10}}+{}^{11}{{C}_{2}}{{\left( x \right)}^{9}}+{}^{11}{{C}_{3}}{{\left( x \right)}^{8}}+......+{}^{11}{{C}_{11}}$.
$\Rightarrow {{\left( 1+x \right)}^{11}}={}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+{}^{11}{{C}_{3}}{{x}^{8}}+......+{}^{11}{{C}_{11}}$ ---(6).
Let us multiply equations (5) and (6).
We get \[\Rightarrow {{\left( 1-x \right)}^{11}}\times {{\left( 1+x \right)}^{11}}=\left( {}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-......-{}^{11}{{C}_{11}}{{x}^{11}} \right)\times \left( {}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+......+{}^{11}{{C}_{11}} \right)\].
\[\Rightarrow {{\left( \left( 1-x \right)\times \left( 1+x \right) \right)}^{11}}=\left( {}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-......-{}^{11}{{C}_{11}}{{x}^{11}} \right)\times \left( {}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+......+{}^{11}{{C}_{11}} \right)\].
\[\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{22}}=\left( {}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-......-{}^{11}{{C}_{11}}{{x}^{11}} \right)\times \left( {}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+......+{}^{11}{{C}_{11}} \right)\].
Now, let us compare the coefficients of ${{x}^{11}}$ on both sides.
Let us find the general form of terms in ${{\left( 1-{{x}^{2}} \right)}^{11}}$. We know that the general form of terms in the binomial expansion ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$, where $0\le r\le n$.
So, the general forms of terms in ${{\left( 1-{{x}^{2}} \right)}^{11}}$ is ${}^{11}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( {{x}^{2}} \right)}^{11-r}}={}^{11}{{C}_{r}}{{\left( x \right)}^{22-2r}}$, $0\le r\le 11$.
We need to find the coefficient of ${{x}^{11}}$. So, we get $22-2r=11$.
$\Rightarrow 2r=11$.
$\Rightarrow r=\dfrac{11}{2}$.
We know that ‘r’ should be integer. So, the coefficient of ${{x}^{11}}$ in ${{\left( 1-{{x}^{2}} \right)}^{11}}$ is 0.
So, we get $0={}^{11}{{C}_{0}}\times {}^{11}{{C}_{0}}+{}^{11}{{C}_{1}}\times \left( -{}^{11}{{C}_{1}} \right)+{}^{11}{{C}_{2}}\times {}^{11}{{C}_{2}}+{}^{11}{{C}_{3}}\times \left( -{}^{11}{{C}_{3}} \right)+......+{}^{11}{{C}_{11}}\times \left( -{}^{11}{{C}_{11}} \right)$.
$\Rightarrow 0={}^{11}C_{0}^{2}-{}^{11}C_{1}^{2}+{}^{11}C_{2}^{3}-{}^{11}C_{3}^{2}+......-{}^{11}C_{11}^{2}$.
From equation (4), we get $z=0$.
So, the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=11$ is 0.
Note: We should know that ‘a’ should be a positive integer and ‘b’ should be a negative integer in the combination ${}^{a}{{C}_{b}}$, which is a very important property. We can see that the given problem contains a huge amount of calculation, so we need to make calculations carefully in every step in order to avoid confusion and mistakes. Whenever we get the problem involving the squares and multiplication of binomial coefficients we should know that there is a possibility of multiplication of two binomial expansions.
Complete step-by-step solution:
According to the problem, we need to find the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=10$ and $n=11$.
Let us first find the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=10$. Let us assume this value is ‘y’.
So, we get $y=~{}^{10}C_{0}^{2}-{}^{10}C_{1}^{2}+{}^{10}C_{2}^{2}-{}^{10}C_{3}^{2}+{}^{10}C_{4}^{2}-{}^{10}C_{5}^{2}+{}^{10}C_{6}^{2}-{}^{10}C_{7}^{2}+{}^{10}C_{8}^{2}-{}^{10}C_{9}^{2}+{}^{10}C_{10}^{2}$ ---(1).
We know that ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{o}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$.
So, we get ${{\left( 1-x \right)}^{10}}={}^{10}{{C}_{o}}+{}^{10}{{C}_{1}}\left( -x \right)+{}^{10}{{C}_{2}}{{\left( -x \right)}^{2}}+{}^{10}{{C}_{3}}{{\left( -x \right)}^{3}}+......+{}^{10}{{C}_{10}}{{\left( -x \right)}^{10}}$.
$\Rightarrow {{\left( 1-x \right)}^{10}}={}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-{}^{10}{{C}_{3}}{{x}^{3}}+......+{}^{10}{{C}_{10}}{{x}^{10}}$ ---(2).
Now, we know that ${{\left( 1+x \right)}^{10}}={{\left( x+1 \right)}^{10}}={}^{10}{{C}_{o}}{{\left( x \right)}^{10}}+{}^{10}{{C}_{1}}{{\left( x \right)}^{9}}+{}^{10}{{C}_{2}}{{\left( x \right)}^{8}}+{}^{10}{{C}_{3}}{{\left( x \right)}^{7}}+......+{}^{10}{{C}_{10}}$.
$\Rightarrow {{\left( 1+x \right)}^{10}}={}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+{}^{10}{{C}_{3}}{{x}^{7}}+......+{}^{10}{{C}_{10}}$ ---(3).
Let us multiply equations (2) and (3).
We get \[\Rightarrow {{\left( 1-x \right)}^{10}}\times {{\left( 1+x \right)}^{10}}=\left( {}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-......+{}^{10}{{C}_{10}}{{x}^{10}} \right)\times \left( {}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+......+{}^{10}{{C}_{10}} \right)\].
\[\Rightarrow {{\left( \left( 1-x \right)\times \left( 1+x \right) \right)}^{10}}=\left( {}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-......+{}^{10}{{C}_{10}}{{x}^{10}} \right)\times \left( {}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+......+{}^{10}{{C}_{10}} \right)\].
\[\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{10}}=\left( {}^{10}{{C}_{o}}-{}^{10}{{C}_{1}}x+{}^{10}{{C}_{2}}{{x}^{2}}-......+{}^{10}{{C}_{10}}{{x}^{10}} \right)\times \left( {}^{10}{{C}_{o}}{{x}^{10}}+{}^{10}{{C}_{1}}{{x}^{9}}+{}^{10}{{C}_{2}}{{x}^{8}}+......+{}^{10}{{C}_{10}} \right)\].
Now, let us compare the coefficients of ${{x}^{10}}$ on both sides.
Let us find the general form of terms in ${{\left( 1-{{x}^{2}} \right)}^{10}}$. We know that the general form of terms in the binomial expansion ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$, where $0\le r\le n$.
So, the general forms of terms in ${{\left( 1-{{x}^{2}} \right)}^{10}}$ is ${}^{10}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( {{x}^{2}} \right)}^{10-r}}={}^{10}{{C}_{r}}{{\left( x \right)}^{20-2r}}$, $0\le r\le 10$.
We need to find the coefficient of ${{x}^{10}}$. So, we get $20-2r=10$.
$\Rightarrow 2r=10$.
$\Rightarrow r=5$.
So, the coefficient of ${{x}^{10}}$ in ${{\left( 1-{{x}^{2}} \right)}^{10}}$ is ${}^{10}{{C}_{5}}$.
So, we get ${}^{10}{{C}_{5}}={}^{10}{{C}_{0}}\times {}^{10}{{C}_{0}}+{}^{10}{{C}_{1}}\times \left( -{}^{10}{{C}_{1}} \right)+{}^{10}{{C}_{2}}\times {}^{10}{{C}_{2}}+{}^{10}{{C}_{3}}\times \left( -{}^{10}{{C}_{3}} \right)+......+{}^{10}{{C}_{10}}\times {}^{10}{{C}_{10}}$.
$\Rightarrow {}^{10}{{C}_{5}}={}^{10}C_{0}^{2}-{}^{10}C_{1}^{2}+{}^{10}C_{2}^{3}-{}^{10}C_{3}^{2}+......+{}^{10}C_{10}^{2}$.
From equation (1), we get $y={}^{10}{{C}_{5}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
$\Rightarrow y=\dfrac{10!}{5!5!}$.
$\Rightarrow y=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}$.
$\Rightarrow y=2\times 9\times 2\times 7$.
$\Rightarrow y=252$.
So, the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=10$ is 252.
Now, let us find the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=11$. Let us assume this value is ‘z’.
So, we get $z=~{}^{11}C_{0}^{2}-{}^{11}C_{1}^{2}+{}^{11}C_{2}^{2}-{}^{11}C_{3}^{2}+{}^{11}C_{4}^{2}-{}^{11}C_{5}^{2}+{}^{11}C_{6}^{2}-{}^{11}C_{7}^{2}+{}^{11}C_{8}^{2}-{}^{11}C_{9}^{2}+{}^{11}C_{10}^{2}-{}^{11}C_{11}^{2}$ ---(4).
Now, we have ${{\left( 1-x \right)}^{11}}={}^{11}{{C}_{o}}+{}^{11}{{C}_{1}}\left( -x \right)+{}^{11}{{C}_{2}}{{\left( -x \right)}^{2}}+{}^{11}{{C}_{3}}{{\left( -x \right)}^{3}}+......+{}^{11}{{C}_{11}}{{\left( -x \right)}^{11}}$.
$\Rightarrow {{\left( 1-x \right)}^{11}}={}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-{}^{11}{{C}_{3}}{{x}^{3}}+......-{}^{11}{{C}_{11}}{{x}^{11}}$ ---(5).
Now, we know that ${{\left( 1+x \right)}^{11}}={{\left( x+1 \right)}^{11}}={}^{11}{{C}_{o}}{{\left( x \right)}^{11}}+{}^{11}{{C}_{1}}{{\left( x \right)}^{10}}+{}^{11}{{C}_{2}}{{\left( x \right)}^{9}}+{}^{11}{{C}_{3}}{{\left( x \right)}^{8}}+......+{}^{11}{{C}_{11}}$.
$\Rightarrow {{\left( 1+x \right)}^{11}}={}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+{}^{11}{{C}_{3}}{{x}^{8}}+......+{}^{11}{{C}_{11}}$ ---(6).
Let us multiply equations (5) and (6).
We get \[\Rightarrow {{\left( 1-x \right)}^{11}}\times {{\left( 1+x \right)}^{11}}=\left( {}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-......-{}^{11}{{C}_{11}}{{x}^{11}} \right)\times \left( {}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+......+{}^{11}{{C}_{11}} \right)\].
\[\Rightarrow {{\left( \left( 1-x \right)\times \left( 1+x \right) \right)}^{11}}=\left( {}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-......-{}^{11}{{C}_{11}}{{x}^{11}} \right)\times \left( {}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+......+{}^{11}{{C}_{11}} \right)\].
\[\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{22}}=\left( {}^{11}{{C}_{o}}-{}^{11}{{C}_{1}}x+{}^{11}{{C}_{2}}{{x}^{2}}-......-{}^{11}{{C}_{11}}{{x}^{11}} \right)\times \left( {}^{11}{{C}_{o}}{{x}^{11}}+{}^{11}{{C}_{1}}{{x}^{10}}+{}^{11}{{C}_{2}}{{x}^{9}}+......+{}^{11}{{C}_{11}} \right)\].
Now, let us compare the coefficients of ${{x}^{11}}$ on both sides.
Let us find the general form of terms in ${{\left( 1-{{x}^{2}} \right)}^{11}}$. We know that the general form of terms in the binomial expansion ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$, where $0\le r\le n$.
So, the general forms of terms in ${{\left( 1-{{x}^{2}} \right)}^{11}}$ is ${}^{11}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( {{x}^{2}} \right)}^{11-r}}={}^{11}{{C}_{r}}{{\left( x \right)}^{22-2r}}$, $0\le r\le 11$.
We need to find the coefficient of ${{x}^{11}}$. So, we get $22-2r=11$.
$\Rightarrow 2r=11$.
$\Rightarrow r=\dfrac{11}{2}$.
We know that ‘r’ should be integer. So, the coefficient of ${{x}^{11}}$ in ${{\left( 1-{{x}^{2}} \right)}^{11}}$ is 0.
So, we get $0={}^{11}{{C}_{0}}\times {}^{11}{{C}_{0}}+{}^{11}{{C}_{1}}\times \left( -{}^{11}{{C}_{1}} \right)+{}^{11}{{C}_{2}}\times {}^{11}{{C}_{2}}+{}^{11}{{C}_{3}}\times \left( -{}^{11}{{C}_{3}} \right)+......+{}^{11}{{C}_{11}}\times \left( -{}^{11}{{C}_{11}} \right)$.
$\Rightarrow 0={}^{11}C_{0}^{2}-{}^{11}C_{1}^{2}+{}^{11}C_{2}^{3}-{}^{11}C_{3}^{2}+......-{}^{11}C_{11}^{2}$.
From equation (4), we get $z=0$.
So, the value of $C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-.....+{{\left( -1 \right)}^{n}}C_{n}^{2}$ for $n=11$ is 0.
Note: We should know that ‘a’ should be a positive integer and ‘b’ should be a negative integer in the combination ${}^{a}{{C}_{b}}$, which is a very important property. We can see that the given problem contains a huge amount of calculation, so we need to make calculations carefully in every step in order to avoid confusion and mistakes. Whenever we get the problem involving the squares and multiplication of binomial coefficients we should know that there is a possibility of multiplication of two binomial expansions.
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