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U elementarnoj algebri, binomni teorem opisuje koeficijente stepena binoma kada je on predstavljen u razvijenoj formi. Njegov najjednostavniji oblik kaže da je
![{\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}\quad \quad \quad (1)}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9EztJCyjlCzgzFaqiNatw0oNFFoAiPyjs5aNi2zti2ztJFoDzDygdD)
za bilo koje realne ili kompleksne brojeve x i y, te bilo koji nenegativan cijeli broj n. Binomni koeficijent, koji se pojavljuje u (1), može se definisati preko funkcije faktorijela n!:
![{\displaystyle {n \choose k}={\frac {n!}{k!\,(n-k)!}}.}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Qoqs2ats5nAi4aNG0aNrEnAiPaNzBnjiNatw2nte3aghBzga0nja5)
Na primjer, pred nama su slučaji kada je 2 ≤ n ≤ 5:
![{\displaystyle (x+y)^{2}=x^{2}+2xy+y^{2}\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Aa2s1aDKOagiNaDK5otiOztdAage2ygaPytw3z2sQzDFDatw3ajJC)
![{\displaystyle (x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9FyjzFoAsOyqhByqo1oDlEz2nBnDzDajG0aNFDnDhDzgiNnDJDoDw2)
![{\displaystyle (x+y)^{4}=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8OzjrFajm3ygoNztm0aNBDoDmPoNs3aNGQaAi5yje1atwOyjFAz2w1)
![{\displaystyle (x+y)^{5}=x^{5}+5x^{4}y+10x^{3}y^{2}+10x^{2}y^{3}+5xy^{4}+y^{5}.\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO85oNo4aNaQoDm2aAeOytoPajFDoNo5yjvCntvEnto4ago1zDwPajhA)
Binomni teorem može se iskazati tako što ćemo reći da je polinomni niz
![{\displaystyle \left\{\,x^{k}:k=0,1,2,\dots \,\right\}\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO83a2nBoDm1ngi4zNKNaNwQoqe0njeNnqvEothBnDKPzjGQotwOzNJB)
iz binomnog tipa.
Jedan način da dokažemo binomni teorem (1) je pomoću matematičke indukcije. Kada je n = 0, imamo da je
![{\displaystyle (a+b)^{0}=1=\sum _{k=0}^{0}{0 \choose k}a^{0-k}b^{k}.}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO82oDFBngoOztw4ztJAnDeQoAwPnjCPnqw3aAwPoNC5nAvFaAnFaAs0)
Sada pretostavimo da teorem važi i kada je eksponent m. Tada, za n = m + 1
![{\displaystyle (a+b)^{m+1}=a(a+b)^{m}+b(a+b)^{m}\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO84ygdAoAzCo2w1aDi5zAdAa2zBzjdFa2sPotKQaDCPyta1ots5nAiO)
![{\displaystyle =a\sum _{k=0}^{m}{m \choose k}a^{m-k}b^{k}+b\sum _{j=0}^{m}{m \choose j}a^{m-j}b^{j}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Azts4zqaOoAzFyta5ytBFnDa5yjG1njw4zjG3otiOo2e2aNeNyqvD)
po hipotezi indukcije
![{\displaystyle =\sum _{k=0}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{j=0}^{m}{m \choose j}a^{m-j}b^{j+1}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO84zqs2zgo5zjiNygvAaDFDnjzAnDKOoNw2aDdBaDm5nAwPnDiOatCO)
množeći sa a i b dobijamo
![{\displaystyle =a^{m+1}+\sum _{k=1}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{j=0}^{m}{m \choose j}a^{m-j}b^{j+1}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO80njm1nqnByteNaqoPo2e2aqwOzta2nAiOoAhCoAa1a2a3zge5aAeN)
izvlačimo član k = 0
![{\displaystyle =a^{m+1}+\sum _{k=1}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{k=1}^{m+1}{m \choose k-1}a^{m-k+1}b^{k}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Dytm0ntC2oDdBngsQnjBBz2o1zNaPntsNnqzAzghAajG2a2wPnjdD)
i kažemo da je j = k − 1
![{\displaystyle =a^{m+1}+\sum _{k=1}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{k=1}^{m}{m \choose k-1}a^{m+1-k}b^{k}+b^{m+1}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Ozqw4ztCQzjBFaDK2zqrFo2vEztlCzDdAyjm3zqa0atC5njlEa2w4)
izvlačimo član k = m + 1 sa desne strane
![{\displaystyle =a^{m+1}+b^{m+1}+\sum _{k=1}^{m}\left[{m \choose k}+{m \choose k-1}\right]a^{m+1-k}b^{k}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8PntoPygi5a2dAoNG0zAzDatFDaNo3nja4nja0aDrAnqeOztBDoDG1)
te kombinujemo dobijene sume
![{\displaystyle =a^{m+1}+b^{m+1}+\sum _{k=1}^{m}{m+1 \choose k}a^{m+1-k}b^{k}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Pagi0zNK5oDrAyqhAz2o5ygrEa2e2yjwPz2iOata2zDo2ztdEnjmN)
iz Pascalovog pravila imamo da je
![{\displaystyle =\sum _{k=0}^{m+1}{m+1 \choose k}a^{m+1-k}b^{k}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Dyti2zqrAntC0zgnFaqa0athAoDi5ytK1ngi2ntzFygrCzjrCzNi2)
dodajemo u m + 1 članova.
Binomni broj je broj u obliku
(kada je n najmanje 2). Kada je znak ili ako je n neparan broj, tada se binomni brojevi kogu rastaviti na faktore algebarski:
![{\displaystyle x^{n}\pm y^{n}=(x\pm y)(x^{n-1}\mp x^{n-2}y+\cdots \mp xy^{n-2}+y^{n-1}).\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Fyta3aqw1njwPzNdDztaNago4oNmQzje1oqvCyjK4ztK3oDs2yti4)
Primjeri:
![{\displaystyle x^{2}-y^{2}=(x-y)(x+y)\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9BoNoPnAi1aqsNoDa2ago5yjFCz2o4atK2aqaOatw1aDw5ote3ytoP)
![{\displaystyle x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9DygsNaNGNztGNaDs2ngwOaNJEoAzCzAdAzAsNzjBFzNKOaDaPaqi2)
![{\displaystyle x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9CzAeQzjrAoqaNoDa2yte4ztCQajGQnDo2oNlFatnAzje5zjw0nqeO)
![{\displaystyle x^{8}-y^{8}=(x-y)(x+y)(x^{2}+y^{2})(x^{4}+y^{4})\,}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO82oNa0zqs3oNC3nAhAzthCajm2nqeQnAe2o2nEaDwQyjG0yjw5ytm4)
Da bi razložili
na faktore, koristite izraz
![{\displaystyle x^{n}-y^{n}=(x-y)\left(\sum _{k=0}^{n-1}x^{k}y^{n-1-k}\right).}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Nz2s1yjzEnjsNnjCNngdAyjC4ygs5aNG2njCNoAsPnjGPaDe5nDnE)
- Amulya Kumar Bag. Binomial Theorem in Ancient India. Indian J.History Sci.,1:68-74,1966.