Binom yoyilmasi

⭐ — imtihonda eng ko'p ishlatiladigan formulalar. · 🟢 — masalani tezlashtiradi · 🟡 — ilg'or daraja · 🔵 — chuqur o'rganish (olimpiada darajasi)

Binom yoyilmasi

• Paskal uchburchagi: har bir son yuqoridagi ikki son yig'indisi. $n$-qator — $(x + y)^n$ koeffitsientlari. Har bir hadda $x$ va $y$ darajalari yig'indisi $n$ ga teng.
• Faktorial: $n! = n \times (n - 1) \times (n - 2) \times \dots \times 1$, $\quad 0! = 1$
• ⭐ Binom koeffitsienti: $\binom{n}{r} = {}^nC_r = \frac{n!}{r!\,(n - r)!}$
• ⭐ $(1 + x)^n$ yoyilmasi: $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots + x^n$
• ⭐ Binom teoremasi ($n$ — musbat butun son): $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$

Binom yoyilmasining umumiy hadi

$(a + b)^n$ ning $(r + 1)$-hadi: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Misol: $(2 + x)^5$ dagi $x^2$ koeffitsienti: $\binom{5}{2} \cdot 2^3 = 10 \cdot 8 = 80$.

Binom koeffitsientlari xossalari

$\binom{n}{0} = \binom{n}{n} = 1 \qquad \binom{n}{1} = n \qquad \binom{n}{r} = \binom{n}{n - r}$

Ikki qavs ko'paytmasidagi koeffitsient

$(p + qx)(1 + x)^n$ dagi $x^k$ koeffitsienti: $p \cdot [x^k \text{ koeffitsienti}] + q \cdot [x^{k-1} \text{ koeffitsienti}]$

Ya'ni ikkinchi qavsdan faqat kerakli darajani beradigan hadlarni olasiz.

🟢 Faktorial va o'rinlashtirishlar

$n! = n(n-1)(n-2)\cdots 2 \cdot 1 \qquad {}^nP_r = \frac{n!}{(n - r)!} \qquad {}^nC_r = \frac{n!}{r!\,(n - r)!}$

🟢 Binom koeffitsientlari xossalari

$\binom{n}{r} + \binom{n}{r + 1} = \binom{n + 1}{r + 1} \quad \text{(Paskal qoidasi)}$

$\sum_{r=0}^{n}\binom{n}{r} = 2^n \qquad \sum_{r=0}^{n}(-1)^r\binom{n}{r} = 0$

• Eng katta koeffitsient o'rtada: $n$ juft bo'lsa $r = \frac{n}{2}$; toq bo'lsa $r = \frac{n - 1}{2}$ va $\frac{n + 1}{2}$.

🟢 $x$ qatnashmagan had (term independent of $x$)

$\left(ax^p + \dfrac{b}{x^q}\right)^n$ ning umumiy hadidagi $x$ darajasi: $p(n - r) - qr$. Uni nolga tenglaysiz:

$r = \frac{pn}{p + q}$

Misol: $\left(x^2 + \dfrac{1}{x}\right)^6$ da $r = \dfrac{2 \cdot 6}{3} = 4$, had: $\binom{6}{4} = 15$.

🔵 Ketma-ket hadlar nisbati

$(1 + x)^n$ da:

$\frac{T_{r+1}}{T_r} = \frac{n - r + 1}{r}\,x$

🟢 Taqribiy hisoblash

$x$ kichik bo'lsa: $(1 + x)^n \approx 1 + nx$

Misol: $1.02^{5} \approx 1 + 5(0.02) = 1.1$ (aniq qiymati $1.104\ldots$).

🟡 Butun bo'lmagan $n$ uchun binom ($|x| < 1$)

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

• $(1 - x)^{-1} = 1 + x + x^2 + x^3 + \dots$ — bu $r = x$ bo'lgan cheksiz G.P.!
• $(1 + x)^{\frac{1}{2}} \approx 1 + \dfrac{x}{2} - \dfrac{x^2}{8}$