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**Hint:**We start solving the by recalling the binomial expansion for the negative exponents as \[{{\left( 1+x \right)}^{-a}}=1+\dfrac{\left( -a \right)}{1}x+\dfrac{\left( -a \right)\left( -a-1 \right)}{2\times 1}{{x}^{2}}+\dfrac{\left( -a \right)\left( -a-1 \right)\left( -a-2 \right)}{3\times 2\times 1}{{x}^{3}}+...+\dfrac{\left( -a \right)\left( -a-1 \right)\left( -a-2 \right)...\left( -a-k+1 \right)}{k\times ...\times 3\times 2\times 1}{{x}^{k}}+...\infty \]. We then find the general term of this expansion and the coefficient of it. We then substitute $a=2$ and $k=n$ to find the coefficient of ${{x}^{n}}$ in the binomial expansion of ${{\left( 1+x \right)}^{-2}}$. We then check what will be the results if n is odd and n is even to get the required result.

**Complete step by step answer:**

According to the problem, we need to find the coefficient of ${{x}^{n}}$ term in the binomial expansion of ${{\left( 1+x \right)}^{-2}}$.

We know that the binomial expansion of ${{\left( 1+x \right)}^{-a}}$ is defined as \[{{\left( 1+x \right)}^{-a}}=1+\dfrac{\left( -a \right)}{1}x+\dfrac{\left( -a \right)\left( -a-1 \right)}{2\times 1}{{x}^{2}}+\dfrac{\left( -a \right)\left( -a-1 \right)\left( -a-2 \right)}{3\times 2\times 1}{{x}^{3}}+...+\dfrac{\left( -a \right)\left( -a-1 \right)\left( -a-2 \right)...\left( -a-k+1 \right)}{k\times ...\times 3\times 2\times 1}{{x}^{k}}+...\infty \] for $\left| x \right|<1$.

We can see that the coefficient of the ${{x}^{k}}$ term is \[\dfrac{\left( -a \right)\left( -a-1 \right)\left( -a-2 \right)...\left( -a-k+1 \right)}{k\times ...\times 3\times 2\times 1}\] ---(1).

Let us compare the expansion ${{\left( 1+x \right)}^{-2}}$ with ${{\left( 1+x \right)}^{-a}}$.

So, we get $a=2$. We substitute this value of a in the equation (1) to find the coefficient of ${{x}^{k}}$ in the binomial expansion ${{\left( 1+x \right)}^{-2}}$.

So, we get the coefficient of the ${{x}^{k}}$ term in the binomial expansion ${{\left( 1+x \right)}^{-2}}$ as \[\dfrac{\left( -2 \right)\left( -2-1 \right)\left( -2-2 \right)...\left( -2-k+1 \right)}{k\times ...\times 3\times 2\times 1}=\dfrac{\left( -2 \right)\left( -3 \right)\left( -4 \right)...\left( -1-k \right)}{k\times ...\times 3\times 2\times 1}={{\left( -1 \right)}^{k}}\dfrac{\left( 2 \right)\left( 3 \right)\left( 4 \right)...\left( k+1 \right)}{k\times ...\times 3\times 2\times 1}\].

Now, we need to find the coefficient of ${{x}^{n}}$ term in the binomial expansion ${{\left( 1+x \right)}^{-2}}$.

Let us assume n is odd. So, we get coefficient as \[{{\left( -1 \right)}^{n}}\dfrac{\left( 2 \right)\left( 3 \right)\left( 4 \right)...\left( n+1 \right)}{n\times ...\times 3\times 2\times 1}=\left( -1 \right)\left( n+1 \right)=-n-1\].

Now, let us assume n is even. We get coefficient as \[{{\left( -1 \right)}^{n}}\dfrac{\left( 2 \right)\left( 3 \right)\left( 4 \right)...\left( n+1 \right)}{n\times ...\times 3\times 2\times 1}=\left( 1 \right)\left( n+1 \right)=n+1\].

We can see that the given options have only one answer which we get only when n is even.

∴ The coefficient of ${{x}^{n}}$ term in the binomial expansion of ${{\left( 1+x \right)}^{-2}}$ is $n+1$.

**So, the correct answer is “Option b”.**

**Note:**We should know that the expansion of ${{\left( 1+x \right)}^{-a}}$ will have infinite terms and is valid only if x satisfies the condition $\left| x \right|<1$. Since the condition $\left| x \right|<1$ is mentioned in the problem, we assume that the value of x lies in it. We should check the cases when the value of n is even and when n is odd while solving this type of problem. Similarly, we can expect problems to find the coefficient of ${{x}^{n}}$ in the binomial expansion ${{\left( 1+x \right)}^{-\dfrac{1}{2}}}$.

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