Question

what is the coefficient of x^0 in ((x+1)^m)*((1+1/(x))^n)

Answer 1

binom(m+n, n)

Answer 2

To find the coefficient of $x^0$ in the given expression, we first simplify the expression:

The given expression is $((x+1)^m)*((1+1/(x))^n)$.

Step 1: Simplify the term $(1+1/x)$. $1 + \frac{1}{x} = \frac{x+1}{x}$

Step 2: Substitute this back into the original expression. $((x+1)^m) * \left(\left(\frac{x+1}{x}\right)^n\right)$ $= (x+1)^m * \frac{(x+1)^n}{x^n}$ $= \frac{(x+1)^{m+n}}{x^n}$ $= x^{-n} (x+1)^{m+n}$

Step 3: Expand $(x+1)^{m+n}$ using the binomial theorem. The binomial theorem states that $(a+b)^N = \sum_{k=0}^{N} \binom{N}{k} a^{N-k} b^k$. Here, $a=x$, $b=1$, and $N=m+n$. So, $(x+1)^{m+n} = \sum_{k=0}^{m+n} \binom{m+n}{k} x^k 1^{m+n-k} = \sum_{k=0}^{m+n} \binom{m+n}{k} x^k$.

Step 4: Multiply the expanded form by $x^{-n}$. $x^{-n} \sum_{k=0}^{m+n} \binom{m+n}{k} x^k = \sum_{k=0}^{m+n} \binom{m+n}{k} x^{k-n}$

Step 5: Identify the term for which the power of $x$ is $0$. We need $x^{k-n} = x^0$, which implies $k-n = 0$. Therefore, $k=n$.

Step 6: Substitute $k=n$ into the general term to find the coefficient. The coefficient of $x^0$ is $\binom{m+n}{n}$.

Alternatively, using the property $\binom{N}{k} = \binom{N}{N-k}$, we can also write $\binom{m+n}{n}$ as $\binom{m+n}{(m+n)-n} = \binom{m+n}{m}$.

The final answer is $\boxed{\binom{m+n}{n}}$.

Explanation of the solution: The expression is simplified to $x^{-n}(x+1)^{m+n}$. The binomial expansion of $(x+1)^{m+n}$ is $\sum_{k=0}^{m+n} \binom{m+n}{k} x^k$. Multiplying by $x^{-n}$, we get $\sum_{k=0}^{m+n} \binom{m+n}{k} x^{k-n}$. For the coefficient of $x^0$, we set the exponent $k-n=0$, which gives $k=n$. Substituting $k=n$ into the binomial coefficient gives $\binom{m+n}{n}$.