题目内容

20.已知数列{an}满足:a1=2,an+1+1=a1a2a3…an
(Ⅰ)求a2的值;
(Ⅱ)(ⅰ)证明:当n≥2时,an2=an+1-an+1;
(ⅱ)若正整数m满足a1a2a3…am+2015=a12+a22+a32+…+am2,求m的值.

分析 (Ⅰ)通过an+1+1=a1a2a3…an,令n=1即得结论;
(Ⅱ)(ⅰ)通过an+1+1=a1a2a3…an及an+1=a1a2a3…an-1可得$\frac{{{a_{n+1}}+1}}{{{a_n}+1}}={a_n}$,进而可得结论;
(ⅱ)通过a1a2a3…am=1+am+1,可得${a_m}^2={a_{m+1}}-{a_m}+1$,利用${a_1}^2+{a_2}^2+{a_3}^2+…+{a_m}^2$=am+1+m+2,计算即可结论.

解答 (Ⅰ)解:∵an+1+1=a1a2a3…an
∴a2+1=a1,∴a2=a1-1=1;
(Ⅱ)(ⅰ)证明:∵an+1+1=a1a2a3…an,①
∴an+1=a1a2a3…an-1,(n≥2).         ②
由①÷②得 $\frac{{{a_{n+1}}+1}}{{{a_n}+1}}={a_n}$,
∴an+1+1=(an+1)an
即当n≥2时${a_n}^2={a_{n+1}}-{a_n}+1$;
(ⅱ)解:由a1a2a3…am=1+am+1
∵${a_1}^2=4$,
${a_2}^2={a_3}-{a_2}+1$,
${a_3}^2={a_4}-{a_3}+1$,

${a_m}^2={a_{m+1}}-{a_m}+1$,
∴${a_1}^2+{a_2}^2+{a_3}^2+…+{a_m}^2$=am+1+m+2,
则(1+am+1)+2015=am+1+m+2,
∴m=2014.

点评 本题考查数列的基本性质,注意解题方法的积累,属于中档题.

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