In expansion $(x^2+1+\frac{1}{x^2})^n$ , n $\in \mathbb{N}$
1.number of terms is $2n+1$
2.coefficient of constant term is $2^{n-1}$
3.coefficient of $x^{2n-2}$ is $n$
4.coefficient of $x^2$ is $n$
I tried by taking lcm and writing as $\frac{(1+x^2+x^4)^n}{x^{2n}}$. How do i proceed? thanks
On
Usign the Newton formula we have:
$\left(x^2+1+\frac{1}{x^2}\right)^n=(x^2+1)^n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
$=x^{2n}+n(x^2)^{n-1}\cdot 1\cdots 1+n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
$=x^{2n}+n(x)^{2n-2}\cdot 1\cdots 1+n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
i.e., concule that 3. is correct.
By using binomial expansion you'll get $$(1+x^2+\frac1{x^2})^n=\sum_{k=0}^{n}{\binom{n}{k}(x^2+\frac1{x^2})^k}$$ Using the theorem one more time gets $$(1+x^2+\frac1{x^2})^n=\sum_{k=0}^{n}{\sum_{j=0}^{k} {\binom{n}{k}\binom{k}{j}(x^2)^{2j-k}}}\label{eq1}\tag{1}$$ Now you can verify the first proposition. For each value of index $k$, expression $2j-k$ produces values from $-k$ to $k$ all with the same parity (odd/even) of $k$. Therefore the greatest value of $k$ determines the total number of terms. Since $k$ is at most equal to $n$ then all powers (of $x^2$) from $-n$ to $n$ are present. So there are $2n+1$ terms. Thus, proposition (1) is true.
Proposition (2) is false. According to (\ref{eq1}), the constant term coefficient is formed by the summation of terms where $2j-k=0$, or $j=k/2$, which implies $k$ should be even. Therefore the constant term coefficient $c_0$ is $$c_0=\sum_{k=0\\ \mathrm{even}}^{n}\binom{n}{k}\binom{k}{k/2}\label{eq2}\tag{2}$$ Obviously $c_0>2^{n-1}$. You can verify this by using the fact that $$\sum_{k=0\\\mathrm{even}}^{n}{\binom{n}{k}}=\sum_{k=1\\\mathrm{odd}}^{n}{\binom{n}{k}}=2^{n-1}$$ which can easily be proven by using the binomial expansion of $(1+x)^n$ for both $x=-1$ and $x=1$, and regrouping odd/even degree terms. Since, for $n\geq2$, each term of the summation in (\ref{eq2}), except the first, is greater than those of the above summation then $c_0>2^{n-1}$.
Proposition (3) is true. The coefficient of $x^{2n-2}$ corresponds to cases where $2j-k=n-1$. Note that since $j\leq k$ then $2j-k\leq k$. So, if $2j-k=n-1$ then $n-1\leq k$. And this means $k$ can only be $n-1$ and $n$. Moreover, $2j-k=n-1$ can be rewritten as $2j=n-1+k$ which cannot be hold when $k=n$ (because of even LHS against odd RHS). Finally $k=n-1$ is the only answer, and hence $j=n-1$, and the coefficient of $x^{2n-2}$, or $c_{2n-2}$, would be $$c_{2n-2}=\binom{n}{n-1}\binom{n-1}{n-1}=n$$
Proposition (4) is false. The coefficient of $x^2$ corresponds to cases where $2j-k=1$, or $2j-1=k$, which means $k$ is odd and for each $k$ only one $j$ is applicable which is $j=(k+1)/2$ (of course for $n>1$). Thus the coefficient of $x^2$, or $c_2$, would be $$c_2=\sum_{k=1\\\mathrm{odd}}^{n}{\binom{n}{k}\binom{k}{(k+1)/2}}$$ With the same argument as in proposition (2), you can conclude that, for $n>2$, $c_2>2^{n-1}$ and hence proposition (4) is false.