Task B-4.1. Solve the equation:

$$
2^{\binom{n+1}{2}}-4 \cdot 2^{\binom{n-1}{2}}=7 \cdot 2^{\binom{n}{2}}
$$
Solution. We solve the equation under the condition that $n \geq 3$.

$$
2^{\frac{(n+1) n}{2}}-4 \cdot 2^{\frac{(n-1)(n-2)}{2}}=7 \cdot 2^{\frac{n(n-1)}{2}}
$$

After dividing the equation by $2^{\frac{n(n-1)}{2}}$ we get

$$
2^{\frac{(n+1) n}{2}-\frac{n(n-1)}{2}}-4 \cdot 2^{\frac{(n-1)(n-2)}{2}-\frac{n(n-1)}{2}}=7
$$

or

$$
2^{n}-4 \cdot 2^{-n+1}=7
$$

If we let $2^{n}=t$ we get the equation $t^{2}-7 t-8=0$ whose solutions are 8 and -1.

Since $2^{n}>0, 2^{n}=8$ so the solution is $n=3$.