Question

If the 17\textsuperscript{th} and the 18\textsuperscript{th} terms in the expansion of $(2 + a)^{50}$ are equal, then the coefficient of $x^{35}$ in the expansion of $(a + x)^{-2}$ is

Answer 1

-36

Answer 2

To solve this problem:

1. Find the value of 'a': Use the equality of the 17th and 18th terms in the binomial expansion of $(2+a)^{50}$. The general term in a binomial expansion is given by $T_{r+1} = \binom{n}{r} X^{n-r} Y^r$.

   • $T_{17} = \binom{50}{16} (2)^{50-16} (a)^{16} = \binom{50}{16} 2^{34} a^{16}$
   • $T_{18} = \binom{50}{17} (2)^{50-17} (a)^{17} = \binom{50}{17} 2^{33} a^{17}$

   Setting $T_{17} = T_{18}$:

   $\binom{50}{16} 2^{34} a^{16} = \binom{50}{17} 2^{33} a^{17}$

   Simplifying, we use the identity $\frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{r+1}{n-r}$, which gives us $a = 1$.

2. Find the coefficient of $x^{35}$ in the expansion of $(a+x)^{-2}$: Substitute $a=1$ into the expression, resulting in $(1+x)^{-2}$. The binomial expansion for $(1+x)^n$ is given by:

   $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$

   In this case, $n=-2$. The coefficient of $x^r$ is given by:

   Coefficient of $x^r = \frac{(-2)(-2-1)(-2-2)\dots(-2-r+1)}{r!} = (-1)^r (r+1)$

   For $x^{35}$, $r=35$, so the coefficient is:

   Coefficient of $x^{35} = (-1)^{35} (35+1) = -36$

Therefore, the coefficient of $x^{35}$ in the expansion of $(a + x)^{-2}$ is -36.