分析:(1)將n=5代入(x+1)n中,變形可得[(x-1)+1]5,則a2為其展開式中(x-1)2的系數(shù),由二項式定理可得答案;
(2)由于與二項式有關(guān),故可采用賦值法.取x=1,則a0=2n,從而可求Sn,再用數(shù)學(xué)歸納法證明即可,只不過需注意假設(shè)n=k時成立,求當(dāng)n=k+1時,Sn增加2k+1-2k=2k項.
解答:解:(1)根據(jù)題意,(x+1)
5=[2+(x-1)]
5=a
0+a
1(x-1)+a
2(x-1)
2+…+a
5(x-1)
5,
則a
2(x-1)
2=C
522
5-2(x-1)
2故a
2=80;
(2)在:(x+1)
n=a
0+a
1(x-1)+a
2(x-1)
2+a
3(x-1)
3+…+a
n(x-1)
n中,
令x=1,可得a
0=2
n,
則
Sn=1+++…+=1
+++…+,
①當(dāng)n=2時,S
n=1
+++…+=1+
+,顯然1<S
n≤2
故當(dāng)n=2時,滿足
<Sn≤n②假設(shè)當(dāng)n=k(k>2,k∈N)時,滿足
<Sn≤n,
即
<S
k=1+
++…+
≤k成立,
當(dāng)n=k+1(k>2,k∈N)時,
S
k+1=1+
++…+
+
+…+
>++…+
>++…+
>+×(
+
…+
)
>+=
S
k+1=1+
++…+
+
+…+
≤k+
+…+
≤k+1
故當(dāng)n=k+1(k>2,k∈N)時,
<S
k+1≤k+1
綜合①②可知,
<Sn≤n,n∈N*,n≥2.
點(diǎn)評:本題主要考查二項式定理的應(yīng)用,注意根據(jù)題意,分析所給代數(shù)式的特點(diǎn),通過給二項式的x賦值,求展開式的系數(shù)和,可以簡便的求出答案,同時考查了數(shù)學(xué)歸納法,屬于中檔題.