Упр.1109 ГДЗ Алимов 10-11 класс (Алгебра)
- Найти значение выражения, предварительно его упростив:
1) $$C_{4}^{1}+C_{4}^{2}+C_{4}^{3}+C_{4}^{4}$$;
2) $$C_{5}^{1}+C_{5}^{2}+C_{5}^{3}+C_{5}^{4}$$;
3) $$C_{7}^{0}+C_{7}^{7}+C_{7}^{1}+C_{7}^{6}+C_{7}^{2}+C_{7}^{5}+C_{7}^{3}+C_{7}^{4}$$;
4) $$C_{6}^{0}+C_{6}^{6}+C_{6}^{1}+C_{6}^{5}+C_{6}^{2}+C_{6}^{4}+C_{6}^{3}$$;
5) $$C_{12}^{2}+C_{12}^{3}+C_{13}^{4}+C_{14}^{5}$$;
6) $$C_{9}^{3}+C_{9}^{4}+C_{10}^{5}+C_{11}^{6}$$;
7) $$C_{21}^{7}-C_{20}^{7}$$;
8) $$C_{25}^{9}-C_{24}^{9}$$;
9) $$C_{15}^{10}-C_{14}^{9}$$;
10) $$C_{13}^{8}-C_{12}^{7}$$.
Используем свойства биномиальных коэффициентов:
$$C_n^k=C_n^{n-k},$$
$$C_n^k+C_n^{k+1}=C_{n+1}^{k+1}.$$
$$C_4^1+C_4^2+C_4^3+C_4^4=C_4^0+C_4^1+C_4^2+C_4^3+C_4^4-C_4^0=2^4-1=15.$$
$$C_5^1+C_5^2+C_5^3+C_5^4=C_5^0+C_5^1+C_5^2+C_5^3+C_5^4-C_5^0-C_5^5=2^5-1-1=30.$$
$$C_7^0+C_7^7+C_7^1+C_7^6+C_7^2+C_7^5+C_7^3+C_7^4=2^7=128.$$
$$C_6^0+C_6^6+C_6^1+C_6^5+C_6^2+C_6^4+C_6^3=2^6=64.$$
$$C_{12}^2+C_{12}^3+C_{13}^4+C_{14}^5=C_{13}^3+C_{13}^4+C_{14}^4+C_{14}^5=C_{14}^4+C_{14}^5=C_{15}^5.$$
$$C_{15}^5=\frac{15!}{5!\,10!}=\frac{15\cdot14\cdot13\cdot12\cdot11}{5\cdot4\cdot3\cdot2}=3003.$$
$$C_9^3+C_9^4+C_{10}^5+C_{11}^6=C_{10}^4+C_{10}^5+C_{11}^6=C_{11}^5+C_{11}^6=C_{12}^6.$$
$$C_{12}^6=\frac{12!}{6!\,6!}=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2}=924.$$
$$C_{21}^7-C_{20}^7=C_{20}^6.$$
$$C_{20}^6=\frac{20!}{6!\,14!}=\frac{20\cdot19\cdot18\cdot17\cdot16\cdot15}{6\cdot5\cdot4\cdot3\cdot2}=38760.$$
$$C_{25}^9-C_{24}^9=C_{24}^8.$$
$$C_{24}^8=\frac{24!}{8!\,16!}=\frac{24\cdot23\cdot22\cdot21\cdot20\cdot19\cdot18\cdot17}{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2}=735471.$$
$$C_{15}^{10}-C_{14}^9=C_{14}^{10}.$$
$$C_{14}^{10}=C_{14}^4=\frac{14!}{4!\,10!}=\frac{14\cdot13\cdot12\cdot11}{4\cdot3\cdot2}=1001.$$
$$C_{13}^8-C_{12}^7=C_{12}^8.$$
$$C_{12}^8=C_{12}^4=\frac{12!}{4!\,8!}=\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2}=495.$$
Ответ: 1) 15; 2) 30; 3) 128; 4) 64; 5) 3003; 6) 924; 7) 38760; 8) 735471; 9) 1001; 10) 495.
