Question

If the number of terms in the expansion of $\left(2+\frac{1}{x}+\frac{1}{x^2}\right)^n$ is 13, then the sum of the coefficients is equal to

• A. 4096

Answer 1

Answer: (A)

4096

Answer 2

To find the sum of the coefficients, we first need to determine the value of $n$.

1. Determine the number of terms in the expansion: The given expression is $\left(2+\frac{1}{x}+\frac{1}{x^2}\right)^n$. Let's consider the general term in the expansion of $(a+b+c)^n$. It is given by $\frac{n!}{r_1! r_2! r_3!} a^{r_1} b^{r_2} c^{r_3}$, where $r_1+r_2+r_3=n$ and $r_1, r_2, r_3 \ge 0$. In this problem, $a=2$, $b=\frac{1}{x}=x^{-1}$, and $c=\frac{1}{x^2}=x^{-2}$. Substituting these values, the general term becomes: $\frac{n!}{r_1! r_2! r_3!} (2)^{r_1} (x^{-1})^{r_2} (x^{-2})^{r_3} = \frac{n!}{r_1! r_2! r_3!} 2^{r_1} x^{-r_2} x^{-2r_3} = \left(\frac{n!}{r_1! r_2! r_3!} 2^{r_1}\right) x^{-(r_2+2r_3)}$ The distinct terms in the expansion depend on the distinct powers of $x$. Let $P = r_2+2r_3$. We need to find the range of possible values for $P$. Since $r_1, r_2, r_3$ are non-negative integers and $r_1+r_2+r_3=n$:

• Minimum value of P: This occurs when $r_2=0$ and $r_3=0$. In this case, $P = 0+2(0)=0$. This corresponds to $r_1=n$, giving the term $2^n x^0$.
• Maximum value of P: This occurs when $r_2$ and $r_3$ are maximized, meaning $r_1$ is minimized (i.e., $r_1=0$). If $r_1=0$, then $r_2+r_3=n$. To maximize $r_2+2r_3$, we should maximize $r_3$. So, set $r_2=0$ and $r_3=n$. In this case, $P = 0+2(n)=2n$. This corresponds to the term $(1/x^2)^n = x^{-2n}$.

Now, we need to check if all integer powers between $0$ and $2n$ can be formed. Let $k$ be an integer such that $0 \le k \le 2n$. We need to show that there exist non-negative integers $r_2, r_3$ such that $r_2+2r_3=k$ and $r_2+r_3 \le n$.

• If $k$ is even: Let $k=2m$. Choose $r_2=0, r_3=m$. Then $r_2+2r_3 = 2m = k$. For this to be valid, we need $r_2+r_3 \le n$, which means $0+m \le n$, or $m \le n$. Since $k \le 2n$, we have $2m \le 2n$, so $m \le n$. Thus, all even powers from $0$ to $2n$ are possible.
• If $k$ is odd: Let $k=2m+1$. Choose $r_2=1, r_3=m$. Then $r_2+2r_3 = 1+2m = k$. For this to be valid, we need $r_2+r_3 \le n$, which means $1+m \le n$. Since $k \le 2n$, we have $2m+1 \le 2n$, which implies $2m \le 2n-1$, so $m \le n-1/2$. Thus $m \le n-1$. Therefore, $1+m \le 1+(n-1) = n$. Thus, all odd powers from $1$ to $2n-1$ are possible.

Combining both cases, all integer powers of $x$ from $x^0$ to $x^{-2n}$ are possible. The number of distinct powers of $x$ is $(2n - 0) + 1 = 2n+1$. Therefore, the number of terms in the expansion is $2n+1$.

We are given that the number of terms is 13. So, $2n+1 = 13$ $2n = 12$ $n = 6$

2. Calculate the sum of the coefficients: To find the sum of the coefficients of a polynomial expression, we substitute all variables with 1. For the expression $\left(2+\frac{1}{x}+\frac{1}{x^2}\right)^n$, the sum of the coefficients is obtained by setting $x=1$. Sum of coefficients $= \left(2+\frac{1}{1}+\frac{1}{1^2}\right)^n$ Sum of coefficients $= (2+1+1)^n$ Sum of coefficients $= (4)^n$

Since we found $n=6$: Sum of coefficients $= 4^6$ $4^6 = (2^2)^6 = 2^{12} = 4096$.