Binomo formulė
Binomo formulė – dažnai dar vadinama Niutono formule, – svarbi matematikos teorema, padedanti rasti dvinario, pakelto n-tuoju laipsniu, skleidinį. Teorema dažniausiai yra užrašoma ( a + b ) n = ∑ k = 0 n ( n k ) a n − k b k = ∑ k = 0 n C n k a n − k b k {\displaystyle (a+b)^{n}=\sum _{k=0}^{n}{n \choose k}a^{n-k}b^{k}=\sum _{k=0}^{n}C_{n}^{k}a^{n-k}b^{k}} arba ( a + b ) n = ( n 0 ) a n + ( n 1 ) a n − 1 b + ⋯ + ( n k ) a n − k b k + ⋯ + ( n n ) b n {\displaystyle (a+b)^{n}={n \choose 0}a^{n}+{n \choose 1}a^{n-1}b+\dots +{n \choose k}a^{n-k}b^{k}+\dots +{n \choose n}b^{n}} Dešinioji šios formulės dalis vadinama binomo laipsnio dėstiniu. Skaičiai ( n k ) = C n k = n ! k ! ⋅ ( n − k ) ! {\displaystyle {n \choose k}=C_{n}^{k}={n! \over {k!\cdot (n-k)!}}} yra vadinami binomo koeficientais ir yra lygūs skaičiams iš atitinkamos Paskalio trikampio eilutės. arba ( a + b ) n = C n 0 a n b 0 + C n 1 a n − 1 b 1 + C n 2 a n − 2 b 2 + ⋯ + C n n − 1 a 1 b n − 1 + C n n a 0 b n {\displaystyle (a+b)^{n}=C_{n}^{0}a^{n}b^{0}+C_{n}^{1}a^{n-1}b^{1}+C_{n}^{2}a^{n-2}b^{2}+\dots +C_{n}^{n-1}a^{1}b^{n-1}+C_{n}^{n}a^{0}b^{n}} kur C n k {\displaystyle C_{n}^{k}} yra deriniai. Jei ( a − b ) n {\displaystyle (a-b)^{n}} , tada bus tai minusas tai pliusas, pradedant nuo minuso, pvz.: ( a − b ) 5 = C 5 0 a 5 − C 5 1 a 4 b + C 5 2 a 3 b 2 − C 5 3 a 2 b 3 + C 5 4 a b 4 − C 5 5 b 5 {\displaystyle (a-b)^{5}=C_{5}^{0}a^{5}-C_{5}^{1}a^{4}b+C_{5}^{2}a^{3}b^{2}-C_{5}^{3}a^{2}b^{3}+C_{5}^{4}ab^{4}-C_{5}^{5}b^{5}} ( a + b ) 2 = a 2 + 2 a b + b 2 {\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}} ( a − b ) 2 = a 2 − 2 a b + b 2 {\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}} ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 {\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}} ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 {\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}} Pastaba, 0 ! = 1. {\displaystyle 0!=1.} Penktos eilės Niutono binomo formulė yra tokia: ( a − b ) 5 = C 5 0 a 5 − C 5 1 a 4 b + C 5 2 a 3 b 2 − C 5 3 a 2 b 3 + C 5 4 a b 4 − C 5 5 b 5 = {\displaystyle (a-b)^{5}=C_{5}^{0}a^{5}-C_{5}^{1}a^{4}b+C_{5}^{2}a^{3}b^{2}-C_{5}^{3}a^{2}b^{3}+C_{5}^{4}ab^{4}-C_{5}^{5}b^{5}=} = 5 ! 0 ! ⋅ ( 5 − 0 ) ! a 5 − 5 ! 1 ! ⋅ ( 5 − 1 ) ! a 4 b + 5 ! 2 ! ⋅ ( 5 − 2 ) ! a 3 b 2 − 5 ! 3 ! ⋅ ( 5 − 3 ) ! a 2 b 3 + 5 ! 4 ! ⋅ ( 5 − 4 ) ! a b 4 − 5 ! 5 ! ⋅ ( 5 − 5 ) ! b 5 = {\displaystyle ={5! \over {0!\cdot (5-0)!}}a^{5}-{5! \over {1!\cdot (5-1)!}}a^{4}b+{5! \over {2!\cdot (5-2)!}}a^{3}b^{2}-{5! \over {3!\cdot (5-3)!}}a^{2}b^{3}+{5! \over {4!\cdot (5-4)!}}ab^{4}-{5! \over {5!\cdot (5-5)!}}b^{5}=} = 5 ! 1 ⋅ 5 ! a 5 − 5 a 4 b + 5 ! 2 ! ⋅ 3 ! a 3 b 2 − 5 ⋅ 4 ⋅ 3 3 ⋅ 2 a 2 b 3 + 5 ! 4 ! a b 4 − 5 ! 5 ! ⋅ 0 ! b 5 = {\displaystyle ={5! \over {1\cdot 5!}}a^{5}-5a^{4}b+{5! \over {2!\cdot 3!}}a^{3}b^{2}-{5\cdot 4\cdot 3 \over {3\cdot 2}}a^{2}b^{3}+{5! \over 4!}ab^{4}-{5! \over {5!\cdot 0!}}b^{5}=} = a 5 − 5 a 4 b + 5 ⋅ 4 2 a 3 b 2 − 5 ⋅ 2 a 2 b 3 + 5 a b 4 − b 5 = {\displaystyle =a^{5}-5a^{4}b+{5\cdot 4 \over 2}a^{3}b^{2}-5\cdot 2a^{2}b^{3}+5ab^{4}-b^{5}=} = a 5 − 5 a 4 b + 10 a 3 b 2 − 10 a 2 b 3 + 5 a b 4 − b 5 . {\displaystyle =a^{5}-5a^{4}b+10a^{3}b^{2}-10a^{2}b^{3}+5ab^{4}-b^{5}.} Užrašysime ketvirto laipsnio Niutono binomo formulę: ( a − b ) 4 = C 4 0 a 4 b 0 − C 4 1 a 3 b + C 4 2 a 2 b 2 − C 4 3 a b 3 + C 4 4 a 0 b 4 = {\displaystyle (a-b)^{4}=C_{4}^{0}a^{4}b^{0}-C_{4}^{1}a^{3}b+C_{4}^{2}a^{2}b^{2}-C_{4}^{3}ab^{3}+C_{4}^{4}a^{0}b^{4}=} = 4 ! 0 ! ( 4 − 0 ) ! a 4 b 0 − 4 ! 1 ! ( 4 − 1 ) ! a 3 b + 4 ! 2 ! ( 4 − 2 ) ! a 2 b 2 − 4 ! 3 ! ( 4 − 3 ) ! a b 3 + 4 ! 4 ! ( 4 − 4 ) ! a 0 b 4 = {\displaystyle ={\frac {4!}{0!(4-0)!}}a^{4}b^{0}-{\frac {4!}{1!(4-1)!}}a^{3}b+{\frac {4!}{2!(4-2)!}}a^{2}b^{2}-{\frac {4!}{3!(4-3)!}}ab^{3}+{\frac {4!}{4!(4-4)!}}a^{0}b^{4}=} = 4 ! 4 ! a 4 − 4 ! 3 ! a 3 b + 4 ! 2 ! ⋅ 2 ! a 2 b 2 − 4 ! 3 ! a b 3 + 4 ! 4 ! b 4 = {\displaystyle ={\frac {4!}{4!}}a^{4}-{\frac {4!}{3!}}a^{3}b+{\frac {4!}{2!\cdot 2!}}a^{2}b^{2}-{\frac {4!}{3!}}ab^{3}+{\frac {4!}{4!}}b^{4}=} = a 4 − 4 a 3 b + 4 ⋅ 3 ⋅ 2 4 a 2 b 2 − 4 a b 3 + b 4 = {\displaystyle =a^{4}-4a^{3}b+{\frac {4\cdot 3\cdot 2}{4}}a^{2}b^{2}-4ab^{3}+b^{4}=} = a 4 − 4 a 3 b + 6 a 2 b 2 − 4 a b 3 + b 4 . {\displaystyle =a^{4}-4a^{3}b+6a^{2}b^{2}-4ab^{3}+b^{4}.}