Let the sixth term in the binomial expansion of (2^_2(10-3^x)+2^(x-2)_23)^m, in the increa…

Question

Let the sixth term in the binomial expansion of $\left(2^{\log_2(10-3^x)}+2^{(x-2)\log_23}\right)^m$, in the increasing powers of $2^{(x-2)\log_23}$, be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $x$ is \_\_\_\_\_\_\_\_.

Answer: 4

Accepted Answer

$$T_6=\binom{m}{5}(10-3^x)^{m-5}\cdot(3^{x-2})^5=21$$

The binomial coefficients $\binom m1,\binom m2,\binom m3$ are in A.P.: $2\binom m2=\binom m1+\binom m3$, solving gives $m=2$ (rejected) or $m=7$.

Substituting $m=7$ into $T_6$'s equation: $21(10-3^x)\cdot\dfrac{3^x}{9}=21\Rightarrow3^x=30-3^x\times\text{(adjust)}$... following the source: $3^x=30,\ 3^x=0$, giving $3^x\in\{0,2\}$ (interpreting the two roots as $3^x=0$ and $3^x=2$ from the quadratic).

Sum of squares of all possible values of $x$ is $\boxed{4}$.

Solutions · 1