## Task B-4.1.

Solve the equation $\binom{x+1}{x-2}+2\binom{x-1}{3}=7(x-1)$.
## First Solution.

Due to the definition of binomial coefficients, it must hold that $x-1 \geqslant 3$, i.e., $x \geqslant 4$.

Using the property of binomial coefficients $\binom{n}{k}=\binom{n}{n-k}$, we can rewrite the given equation as:

$\binom{x+1}{x+1-x+2}+2\binom{x-1}{3}=7(x-1)$, or

$\binom{x+1}{3}+2\binom{x-1}{3}=7(x-1)$.

1 point

By applying the rule for calculating binomial coefficients, we get

$\frac{(x+1) x(x-1)}{6}+2 \frac{(x-1)(x-2)(x-3)}{6}=7(x-1)$, and after dividing by $(x-1) \neq 0$, we have

$\frac{(x+1) x}{6}+2 \frac{(x-2)(x-3)}{6}=7$

This equation reduces to the quadratic equation $x^{2}-3 x-10=0$, 1 point

whose solutions are $x_{1}=5, x_{2}=-2$. 1 point

Due to the initial conditions, the solution to the given equation is $x=5$. 1 point