[考綱傳真] 會(huì)用二項(xiàng)式定理解決與二項(xiàng)展開(kāi)式有關(guān)的簡(jiǎn)單問(wèn)題.
1.二項(xiàng)式定理
(1)二項(xiàng)式定理:(a+b)n=Can+Can-1b+…+Can-rbr+…+Cbn(n∈N*);
(2)通項(xiàng)公式:Tr+1=Can-rbr,它表示第r+1項(xiàng);
(3)二項(xiàng)式系數(shù):二項(xiàng)展開(kāi)式中各項(xiàng)的系數(shù)C,C,…,C.
2.二項(xiàng)式系數(shù)的性質(zhì)
(1)0≤r≤n時(shí),C與C的關(guān)系是.
(2)二項(xiàng)式系數(shù)先增大后減中間項(xiàng)最大
當(dāng)n為偶數(shù)時(shí),第-1項(xiàng)的二項(xiàng)式系數(shù)最大,最大值為Cn;當(dāng)n為奇數(shù)時(shí),第項(xiàng)和項(xiàng)的二項(xiàng)式系數(shù)最大,最大值為
1.C+C+C+…+C=2n.
2.C+C+C+…=C+C+C+…=2n-1.
[基礎(chǔ)自測(cè)]
1.(思考辨析)判斷下列結(jié)論的正誤.(正確的打“√”,錯(cuò)誤的打“×”)
(1)Can-kbk是(a+b)n的展開(kāi)式中的第k項(xiàng). ( )
(2)二項(xiàng)展開(kāi)式中,系數(shù)最大的項(xiàng)為中間一項(xiàng)或中間兩項(xiàng). ( )
(3)(a+b)n的展開(kāi)式中某一項(xiàng)的二項(xiàng)式系數(shù)與a,b無(wú)關(guān). ( )
(4)通項(xiàng)Tk+1=Can-kbk中的a和b不能互換. ( )
[答案] (1)× (2)× (3)√ (4)√
2.(教材改編)(1-2x)4展開(kāi)式中第3項(xiàng)的二項(xiàng)式系數(shù)為( )
A.6 B.-6
C.24 D.-24
A [(1-2x)4展開(kāi)式中第3項(xiàng)的二項(xiàng)式系數(shù)為C=6.故選A.]